MATHEMATICS - SprueMaster

MATHEMATICS 

FOR THE 

AVIATION TRADES 

by JAMES NAIDICH 

Chairman, Department of Mafhe mati r.v, 

Manhattan High School of Aviation Trades 

MrGKAW-IIILL HOOK COMPANY, INC. 

N JO W Y O K K AND LONDON

MATHEMATICS FOR THK AVI VTION TRADES 

COPYRIGHT, 19I2, BY THK 

BOOK TOMPVNY, INC. 

PRINTED IX THE UNITED STATES OF AMERICA 

AIL rights referred. Tin a book, or 

parts thereof, may not be reproduced 

in any form without perm 'nation of 

the publishers.

PREFACE 

This book has been written for students in trade and 

technical schools who intend to become aviation mechanics. 

The text has been planned to satisfy the demand on the 

part of instructors and employers that mechanics engaged 

in precision work have a thorough knowledge of the fundamentals 

of arithmetic applied to their trade. No mechanic 

can work intelligently from blueprints or use measuring 

tools, such as the steel rule or micrometer, without a knowledge 

of these fundamentals. 

Each new topic 

is 

presented as a job, thus stressing the 

practical aspect of the text. Most jobs can be covered in 

one lesson. However, the interests and ability of the group 

will in the last analysis determine the rate of progress. 

Part I is entitled "A Review of Fundamentals for the 

Airplane Mechanic." The author has found through actual 

experience that mechanics and trade-school students often 

have an inadequate knowledge of a great many of the points 

covered in this part of the book. This review will serve to 

consolidate the student's information, to reteach what he 

may have forgotten, to review what he knows, and to 

provide drill in order to establish firmly the basic essentials. 

Fractions, decimals, perimeter, area, angles, construction, 

and graphic representation are covered rapidly but 

systematically. 

For the work in this section two tools are needed. First, 

a steel rule graduated in thirty-seconds and sixty-fourths 

is indispensable. It is advisable to have, in addition, an 

ordinary ruler graduated in eighths and sixteenths. Second, 

measurement of angles makes a protractor necessary.

vi 

Preface 

Parts II, III, and IV deal with specific aspects of the work 

that an aviation mechanic may encounter. The airplane 

and its wing, the strength of aircraft materials, and the mathematics 

associated with the aircraft engine are treated as 

separate units. All the mathematical background required 

for this work is covered in the first part of the book. 

Part V contains 100 review examples taken from airplane 

shop blueprints, aircraft-engine instruction booklets, airplane 

supply catalogues, aircraft directories, and other 

trade literature. The airplane and its engine are treated 

as a unit, and various items learned in other parts of the 

text are coordinated here. 

Related trade information is closely interwoven with the 

mathematics involved. Throughout the text real aircraft 

data are used. Wherever possible, photographs and tracings 

of the airplanes mentioned are shown so that the student 

realizes he is dealing with subject matter valuable not only 

as drill but worth remembering as trade information in his 

elected vocation. 

This book obviously does not present all the mathematics 

required by future aeronautical engineers. All mathematical 

material which could not be adequately handled by 

elementary arithmetic was omitted. The author believes, 

however, that the student who masters the material 

included in this text will have a solid foundation of the type 

of mathematics needed by the aviation mechanic. 

Grateful acknowledgment is made to Elliot V. Noska, 

principal of the Manhattan High School of Aviation Trades 

for his encouragement and many constructive suggestions, 

and to the members of the faculty for their assistance in 

the preparation of this text. The author is also especially 

indebted to Aviation magazine for permission to use 

numerous photographs of airplanes and airplane parts 

throughout the text. 

JAMES NAIDICH. 

NEW YORK.

CONTENTS 

PREFACE 

FOREWORD BY ELLIOT V. NOSKA 

PAOH 

v 

ix 

PART I 

A REVIEW OF FUNDAMENTALS FOR THE AIRPLANE 

OH 

MECHANIC 

I. THE STEEL RULE 3 

II. DECIMALS IN AVIATION 20 

III. MEASURING LENGTH 37 

IV. THE AREA OF SIMPLE FIGURES 47 

V. VOLUME AND WEIGHT 70 

VI. ANGLES AND CONSTRUCTION 80 

VII. GRAPHIC REPRESENTATION OF AIRPLANE DATA ... 98 

PART II 

THE AIRPLANE AND ITS WING 

VIII. THE WEIGHT OF THE AIRPLANE 113 

IX. AIRFOILS AND WING RIBS 130 

PART III 

MATHEMATICS OF MATERIALS 

X. STRENGTH OF MATERIAL 153 

XL FITTINGS, TUBING, AND RIVETS 168 

XII. BEND ALLOWANCE 181 

PART IV 

AIRCRAFT ENGINE MATHEMATICS 

XIII. HORSEPOWER 193 

XIV. FUEL AND OIL CONSUMPTION 212 

XV. COMPRESSION RATIO AND VALVE TIMING 224 

PART V 

REVIEW 

XVI. ONE HUNDRED SELECTED REVIEW EXAMPLES. . 

. . 241 

APPENDIX: TABLES AND FORMULAS 259 

INDEX 265 

vii

FOREWORD 

Aviation is fascinating. Our young men and our young 

women will never lose their enthusiasm for wanting to 

know more and more about the world's fastest growing 

and most rapidly changing industry. 

We are an air-conscious nation. Local, state, and federal 

agencies have joined industry in the vocational training 

of our youth. This is the best guarantee of America's 

continued progress in the air. 

Yes, aviation is 

fascinating in its every phase, but it is 

not all glamour. Behind the glamour stands the training 

and work of the engineer, the draftsman, the research 

worker, the inspector, the pilot, and most important of all, 

the training and hard work of the aviation mechanic. 

Public and private schools, army and navy training 

centers have contributed greatly to the national defense 

effort by training and graduating thousands of aviation 

mechanics. These young men have found their place in 

airplane factories, in approved repair stations, and with 

the air lines throughout the country. 

The material in Mathematics for the Aviation Trades 

has been gathered over a period of years. It has been 

tried out in the classroom and in the shop. For the instructor, 

it solves the problem of what to teach and how to 

teach it. The author has presented to the student mechanic 

in the aviation trades, the necessary mathematics which 

will help him while receiving his training 

in the school 

shop, while studying at home on his own, and while actually 

performing his work in industry. 

The mechanic who is seeking advancement will find here 

a broad background of principles of mathematics relating 

to his trade. 

IX

x 

Foreword 

The text therefore fills a real need. I firmly believe that 

the use of this book will help solve some of the aviation 

mechanic's problems. It will help him to do his work 

more intelligently and will enable him to progress toward 

the goal he has set for himself. 

ELLIOT V. 

NOSKA, 

NEW YORK, Principal, Manhattan High 

December, 1941. School of Aviation Trades

Part 

I 

A REVIEW OF FUNDAMENTALS FOR THE 

AIRPLANE MECHANIC 

Chapter I: 

The Steel Rule 

Job 1 : 

Learning to Use the Rule 

Job 2: Accuracy of Measurement 

Job 3 : 

Reducing Fractions to Lowest Terms 

Job 4: An Important Word: Mixed Number 

Job 5: Addition of Ruler Fractions 

Job 6: Subtraction of Ruler Fractions 

Job 7: Multiplication of Fractions 

Job 8: Division of Fractions 

Job 9: Review Test 

Chapter II: Decimals in Aviation 

Job 1 : 

Reading Decimals 

Job 2: Checking Dimensions with Decimals 

Job 3: Multiplication of Decimals 

Job 4: Division of Decimals 

Job 5: Changing Fractions to Decimals 

Job 6: The Decimals Equivalent Chart 

Job 7: Tolerance and Limits 


Chapter III: Measuring Length 

Job 1 : Units of Length 

Job 2: Perimeter 

Job 8: Xon-ruler Fractions 

Job 4: The Circumference of a Circle 


Chapter IV: The Area of Simple Figures 

Job 1 : Units of Area 

Job 2: The Rectangle 

Job 3: Mathematical Shorthand: Squaring a Number 

1

2 Mathematics for the Aviation Trades 

Job 4: Introduction to Square Roots 

Job 5: The Square Root of a Whole Number 

Job 6: The Square Root of Decimals 

Job 7 : The Square 

Job 8: The Circle 

Job 9: The Triangle 

Job 10: The Trapezoid 


Chapter V: Volume and Weight 

Job 1 : Units of Volume 

Job 2: The Formula for Volume 

Job 3: The Weight of Materials 

Job 4: Board Feet 


Chapter VI: Ansles and Constructions 

Job 1: How to Use the Protractor 

Job 2: How to Draw an Angle 

Job 3: Units of Angle Measure 

Job 4: Angles in Aviation 

Job 5 : To Bisect an Angle 

Job 6: To Bisect a Line 

Job 7: To Construct a Perpendicular 

Job 8: To Draw an Angle Equal to a Given Angle 

Job 9: To Draw a Line Parallel to a Given Line 

Job 10: To Divide a Line into Any Number of Equal Parts 


Chapter VII: Graphic Representation of Airplane Data 

Job l:The Bar Graph 

Job 2: Pictographs 

Job 3: The Broken-line Graph 

Job 4: The Curved-line Graph 

Job 5: Review Test

I 

I 

Chapter I 

THE STEEL RULE 

Since the steel rule is one of the most useful of a mechanic's 

tools, it is 

very important 

it 

quickly and accurately. 

Job 1 : 

Learning to Use the Rule 

for him to learn to use 

Skill in using the rule depends almost entirely on the 

amount of practice obtained in measuring and in drawing 

lines of a definite length. The purpose of this job is to give 

the student some very simple practice work and to stress 

the idea that accuracy of measurement is essential. There 

should be no guesswork on any job; there must be no guesswork 

in aviation. 

Fig. 1 . Steel rule. 

In Fig. 1 is a diagram of a steel rule graduated in 3 c 2nds 

and G4ths. The graduations (divisions of each inch) are 

extremely close together, but the aircraft mechanic is 

often expected to work to the nearest 64th or closer. 

Examples: 

1. How are the rules in Figs. 2a and 26 graduated? 

T M 

Fig. 2a. Fig. 2b.

' 

Mathematics for the Aviation Trades 

2. Draw an enlarged diagram of the first inch of a steel 

rule graduated in 8ths. Label each graduation. 

3. Draw an enlarged diagram of 1 in. of a rule graduated 

(a) in 16ths, (6) in 32nds, (c) in 64ths. 

4. See how quickly you can name each of the graduations 

indicated by letters A, B, C, etc., in Fig. 3. 

IPPP|1P|^^ 

95 8fr Ofr K *l 91 9 I 

9S fit' Of ?C frZ 91 9 I 9S 8V 0* ZC VI 91 8 

| 

9S 91- Ofr ZC W 91 8 

> z 3 

4 8 I? 16 20 24 28 I 4 8 I? 16 20 24 28 48 12 16 20 24 2fl 48 

I? 16 20 24 28 

ililililililililililililihlililililililililililililililililili ililihlililililililililililili ililihlililililililililililih 

D E F 

Fig. 3. // 7 

6. Measure the length of each of the lines in Fig. 4, using 

a rule graduated in 32nds. 

-\ : h H TTl \- 

(CL) (b) (c) 

(cL) 

(e) 

(f) 

Fig. 4. 

H- -H- 

6. Carefully draw the following lengths: 

(a) I in. (6) fV in. (c) & in. (d) %V in. (e) 1^ in. 

7. Measure each of the dimensions in Fig. 5. Read the 

< F -----H 

Fig. 5. Top view of an airplane.

The Steel Rule 

nearest graduation (a) using a rule graduated in 16ths, 

(b) using a rule graduated in 64ths. 

8. Estimate the length of the lines in Fig. 6; then measure 

them with a rule graduated in 64ths. See how well you can 

judge the length 

of a line. 

Write the answers in your own notebook. Do NOT write in 

your textbook. s. 6. 

Job 2: Accuracy of Measurement 

Many 

mechanics find it difficult to understand that 

nothing can ever be measured exactly. For instance, a 

piece of metal is measured with three different rules, as 

Fi 9 . 7. 

shown in Fig. 7. 

Notice that there is a considerable difference 

in the answers for the length, when measured to the 

nearest graduation. 

1 . The rule graduated in 4ths gives the answer f in.


2. The rule graduated in 8ths gives the answer |- in. 

3. The rule graduated in IGths give the answer y-f in. 

Since the first rule is graduated in 4ths, it can be used to 

measure to the nearest quarter of an inch. Therefore, f- in. 

1 

is the correct answer for the length to the nearest quarter. 

The second rule measures to the nearest 8th (because it is 

graduated in Hths) and |- in. is the correct answer to the 

nearest 8th of an inch. Similarly, the answer y|- in. is correct 

to the nearest I6th. If it were required to measure to the 

nearest 32nd, none of these answers would be correct, 

because a rule graduated in 32nds would be required. 

What rule would be required to measure to the nearest 

64th of an inch? 

To obtain the exact length of the metal shown in the 

figure, a rule (or other measuring instrument) with an 

infinite number of graduations per inch would be needed. 

No such rule can be made. No such rule could be read. The 

micrometer can be used to measure to the nearest thousandth 

or ten-thousandth of an inch. Although special devices 

can be used to measure almost to the nearest millionth 

of an inch, not even these give more than a very, very, close 

approximation of the exact measurement. 

The mechanic, therefore, should learn the degree of 

accuracy required for each job in order to know how to 

make and measure his work. This information is 

generally 

given in blueprints. Sometimes it is left to the best judgment 

of the mechanic. Time, price, purpose of the job, and 

measuring tools available should be considered. 

The mechanic who carefully works to a greater than 

necessary degree of accuracy is wasting time and money. 

The mechanic who carelessly works to a degree of accuracy 

less than that which the job requires, often wastes material, 

time, and money. 

1 

When a line is measured by reading the nearest ruler graduation, the possible 

error cannot be greater than half the interval between graduations. Thus $ in. 

is the correct length within J in. See ('hap. II, Job 7, for further information 

on accuracy of measurements.

the Steel Rule 

Examples: 

1. What kind of rule would you use to measure (a) to the 

nearest 16th? (6) 

to the nearest 32nd? 

2. Does it make any difference whether a mechanic works 

to the nearest 16th or to the nearest 64th? Give reasons for 

your answer. 

3. To what degree of accuracy is work generally done in 

(a) a woodworking shop? (6) a sheet metal shop? (c) a 

machine shop? 

4. Measure the distance between the points in Fig. 8 to 

the indicated degree of accuracy. 

Note: A point 

is indicated by the intersection of two lines 

as shown in the figure. What students sometimes call a 

more correctly known as a blot. 

point is 

Fig. 8. 

5. In aeronautics, the airfoil section is the outline of the 

wing rib of an airplane. Measure the thickness of the airfoil 

section at each station in Fig. 9, to the nearest 64th. 

Station 1 

Fig. 9. 

Airfoil section.


6. What is the distance between station 5 and station 9 

(Fig. 9)? 

7. How well can you estimate the length of the lines in 

10? Write down your estimate in your own notebook; 

Fig. 

then measure each line to the nearest 32nd. 

H 

K 

Fi g . 10. 

8. In your notebook, try to place two points exactly 

1 in. apart without using your rule. Now measure the distance 

between the points. How close to an inch did you 

come ? 

Job 3: Reducing Fractions to Lowest Terms 

A. Two Important Words: Numerator, Denominator. 

You probably know that your ruler is graduated in fractions 

or parts of an inch, such as f , ^, j/V, etc. Name any other 

fractions found on it. Name 5 fractions not found on it. 

These fractions consist of two parts separated by a bar or 

fraction line. Remember these two words: 

Numerator is the number above the fraction line. 

Denominator is the number below the fraction line. 

For example, in the fraction |- , 

5 is the numerator and 8 

is the denominator. 

Examples: 

1. Name the numerator and the denominator in each of 

these fractions : 

f 7 16 5 

8"> ~3~> > 

13 

T8~> 

1 

16

, 

The Steel Rule 9 

2. Name 5 fractions in which the numerator is smaller 

than the denominator. 

3. Name 5 fractions in which the numerator is larger than 

the denominator. 

4. If the numerator of a fraction is equal to the denominator, 

what is 

the value of the fraction? 

5. What part of the fraction -g^- shows that the measurement 

was probably made to the nearest 64th? 

B. Fractions Have Many Names. It may have been 

noticed that it is 

possible to call the same graduation on a 

rule bv several different names. _, ... 

, 

This graduation 

.. 

con be ecu Iled 

Students sometimes ask, $ or $ or jj or j$ , etc* 

" which of these ways of calling 

a fraction is moat correct?" All 

of them are "equally" correct. * 

t 

However, it is very useful to be 

IS 

able to change a fraction into 

' 

an equivalent fraction with a different numerator and 

denominator. 

Examples: 

Answer these questions with the help of Fig. 11: 

3 _ how many? 9 ~ -~ 

h w many? 

32 

1- 4 

= 

HT^" 4 

qoL o._l_ 4 - = 2 

d - 

3 ? 3 ? 

^8 ~~ ^1() 8 *32 

Hint: Multiplying the numerator and denominator of 

any fraction by the same number will not change the value 

of the fraction. 

~ 3 2. 

2 4 8 16 

K i_ ? fi 

- 

7 A__L ~ 

8 1 =-1 = -^ 

' 

.'52 

(>4 8 1 32 

e V ? 1 ? 

Q^_' _ IA a i 9 1 

9> 8 

~ 

Te 

~ 

4 

10> * 2 

- * 4


11

I 

' 

I 

' 

I 

! 


Write 5 whole numbers. Write 5 fractions. Write 5 

mixed numbers. 

Is this statement true: Every graduation on a rule, 

beyond the 1-in. mark, corresponds to a mixed number? 

Find the fraction f on a rule? 

-XT j." ! . . i i j.i_ 

The fraction / is {he same 

Notice that it is beyond the 

8 

asfhemfxed number ^ 

1-in. 

1 

graduation, and by actual 

I 

count is equal to 1-g- in. 

A. t 

Changing Improper Fractions 

to Mixed Numbers. Any 

improper fraction (numerator 

larger than the denominator) can be changed to a mixed 

number by dividing the numerator by the denominator. 

Examples: 

ILLUSTRATIVE EXAMPLE 

Change | to a mixed number. 

J = 9 -5- 8 = IS Ans. 

Change these fractions to mixed numbers: 

1 -r- 2 3. 4. 5. ^-f 

6*. 

V ?! |f 

8.' |f 

9.' 

W 10.' 

W- 

11. ^ 12. ^ 13. f| 14. ff 15. ff 

16. Can all fractions be changed 

to mixed numbers? 

Explain. 

B. Changing Mixed Numbers to Improper Fractions. 

A mixed number may be changed to a fraction. 

Change 2| 


to a fraction. 

44 44 

Check your answer by changing the fraction back to a mixed 

number.

1 2 Mathematics for the Aviation Trades 

Examples: 

Change the following mixed numbers to improper fractions. 

Check your answers. 

1. 31 2. 4| 3. 3f 4. 6| 

5. lOf 6. 12| 7. 19^ 8. 2ff 

Change the following improper fractions to mixed 

numbers. Reduce the answers to lowest terms. 

Q 9 

10 JLL 11 45 19 

2 A 

V. 2 *v. 16 4 * 04 

1 Q 5 4 A 17 1 K 4 3 1 C 3 5 

Id. 3-2 14. IT 10. rs 16. r(r 

Job 5: Addition of Ruler Fractions 

A mechanic must work with blueprints or shop drawings, 

which at times look something like the diagram in Fig. IS. 

t 

/---< 

7 " 

Overall length 

------------------------ -H 

Fis. 13. 


Find the over-all length of the job in Fig. 13. 

Over-all length = 1& + J + 

Sum = 4f| 

Method: 

a. Give all fractions the same denominator. 

b. Add all numerators. 

c. Add all whole numbers. 

d. Reduce to lowest terms. 

Even if the "over-all length" is given, it is up 

to the 

intelligent mechanic to check the numbers before he goes 

ahead with his work.


Examples: 

Add these ruler fractions: 

1. 1- + i 

3. I- + !- 

5. I + -I + 

7. i + 1 + i 

a.i+i + 

4- f + 1 + 

6. I + I + 

8. Find the over-all length of the diagram in Fig. 14: 

Add the following: 

J 

Overall length--- 

>l 

Fig. 14. Airplane wing rib. 

9. 1\ + 1| 

10. 3| + 2i 

11. 41- + 2TV 

12. 2i- + If- + 11 

13. 3^ + 5-/ t + 9 TV 

14. 3 + 4 

15. ;J^ + 4 + ^8- 

Fig. 15. 

16. Find the over-all dimensions of the fitting shown in 

Fig. 15. 

Job 6: Subtraction of Ruler Fractions 

The subtraction of ruler fractions is useful in finding 

missing dimensions on blueprints and working drawings. 


In Fig. 16 one of the important dimensions was omitted. 

Find it and check your answer.

14 Mat/temat/cs for the Aviation Trades 

3f = S| 

If in. 

Ans. 

Check: Over-all length = If + 2f = 3. 

t 

Fig. 16. 

Examples: 

Subtract these fractions : 

rj 

9 4I- 1 3 

^ ^4 ~" 8 

5ol 

5 

T/ie Steel Rule 15 

Job 7: Multiplication of Fractions 

The multiplication of fractions has many very important 

applications and is almost as easy as multiplication of whole 

numbers. 

ILLUSTRATIVE EXAMPLES 

35 3X5 15 

Multiply 2 X |. 

VX 

88 8~X~8 04 



n 

' " 

Take } 

of IS. 

1 15 _ 1 X 15 _ 15 

4 X 8 4X8" 3 

' 

Method: 

a. Multiply the numerators, then the denominators. 

b. 

Change all mixed numbers to fractions first, if 

necessary. 

Cancellation can often be used to make the job of multiplication 

easier. 

Examples: 

1. 4 X 3| 2. -?- 

of 8;V 3. 16 X 3f 

4. of 18 5. V X -f X M 6. 4 X 2i X 1 

7. 33 X I X \ 

8. Find the total length of 12 pieces of round stock, each 

7$ in. long. 

9. An Airplane rib weighs 1 jf Ib. What is the total weight 

of 24 ribs? 

10. The fuel tanks of the Bellanca Cruisair hold 22 gal. 

of gasoline. What would it cost to fill this tank at 25^


34 f ~2" 

Fig. 19. Bellanca Cruisair low-wins monoplane. (Courtesy of Aviation.) 

Job 8: Division of Fractions 

A. Division by Whole Numbers. Suppose that, while 

working on some job, a mechanic had to shear the piece of 

metal shown in Fig. 20 into 4 equal parts. The easiest way of 

doing this would be to divide J) T by 4, and then mark the 

points with the help of a rule. 

4- -h 

Fig. 20. 


Divide 9j by 4. 

91- + 4 = 91 X | 

= -V X t = H = 2& Ans. 

Method: 

To divide any fraction by a whole number, multiply by 1 

the whole number. 

over 

Examples: 

How quickly can you get the correct answer? 

1. 4i + 3 

2. 2f -r 4 

4. f H- S 5. 4-5-5 

3. 7| -s- 9 

6. A + 6


7. The metal strip in Fig. 21 is to be divided into 4 equal 

parts. Find the missing dimensions. 

7 " 

3% 

+ 

g. 21. 

8. Find the wall thickness of the tubes in Fig. 22. 

Fig. 22. 

B. Division by Other Fractions. 

Method: 


Divide 3f by |. 

3g - 1 = S| X | 

= ^ X | 

= 

Complete this example. 

How can the answer be checked? 

To divide any fraction by a fraction, invert the second fraction 

and multiply. 

Examples: 

1. I - I 2. li - | 

4. 12f - i 5. 14f - If 

3. Of -5- | 

6. l(>i - 7f 

7. A pile of aircraft plywood is 7^ in. high. Each piece 

is i% in. thick. How many pieces are there altogether? 

8. A piece of round stock 12f in. long is to be cut into 

8 equal pieces allowing Y$ in. for each cut. What is the


length of each piece? Can you use a steel rule to measure 

this distance? Why? 

9. How many pieces of streamline tubing each 4-f in. 

long can be cut from a 72-in. length? Allow ^2 in. for each 

cut. What is the length of the last piece? 

10. Find the distance between centers of the equally 

spaced holes in Fig. 23. Fi3 . 23. 


1. Find the over-all lengths' in Fig. 24. .* 

'64 

2. Find the missing dimensions in Fig. 25. 

(a) 

Fig. 25. 

3. One of the dimensions of Fig. 

Can you find it? 

26 has been omitted.


u 

Fig. 26. 

4. What is the length of A in Fig. 27 ? 

-p 

Fig. 27. Plumb bob. 

of 70 

5. The Curtiss-Wright A-19-R has fuel capacity 

gal. and at cruising speed uses 29f gal. per hour. How many 

hours can the plane stay aloft ? 

Fig. 28. 

Curtiss-Wright A-19-R. (Courtesy of Aviation.) 

6. How well can you estimate length? Check your estimates 

by measuring to the nearest 32nd (see Fig. 29). 

A+ +B 

+D 

Fig. 29.

Chapter II 

DECIMALS IN 

AVIATION 

The ruler is an excellent tool for measuring the length 

of most things but its accuracy is limited to -$% in. or less. 

For jobs requiring a high degree of accuracy the micrometer 

caliper should be used, because it measures to the nearest 

thousandth of an inch or closer. 

Spindle, 

HmlniE 

Thimble 

Sleeve 

Frame 

Job 1 : 

Fig. 30. Micrometer caliper. 

Reading Decimals 

When a rule is used to measure length, the answer is 

expressed as a ruler fraction, such as |, 3^V, or 5^. When 

a micrometer is used to measure length, the answer is 

expressed as a decimal fraction. A decimal fraction is a 

special kind of fraction whose denominator is either 10, 100, 

1,000, etc. For example, yV is a decimal fraction; so are 

T^ and 175/1,000. 

For convenience, these special fractions are written in 

this way: 

10 

= 0.7, read as seven tenths 

20

Decimals in Aviatior? 21 

~ 

35 

= 0.35, read as thirty-five hundredths 

5 

1,000 

45 

10,000 

- = 0.005 read as five thousandths 

= 0.0045, read as forty-five ten-thousandths, 

or four and one-half thousandths 

Examples: 

1. Read these decimals: 

2. Write these decimals: 

(a) 45 hundredths (b) five thousandths 

(e) 3 and 6 tenths (rf) seventy-five thousandths 

(e) 35 ten-thousandths (/) one and three thousandths 

Most mechanics will not find much use for decimals 

beyond the nearest thousandth. When a decimal is given in 

(> 

places, as in the table of decimal equivalents, not all of 

these places should or even can be used. The type of work 

the mechanic is doing will determine the degree of accuracy 

required. 

Method: 


Express 3.72648: 

(a) to the nearest thousandth 3.7 C 2(> Ans. 

(b) 

(c) 

to the nearest hundredth 3.73 Ans. 

to the nearest tenth 3.7 Aus. 

a. Decide how many decimal places your answer should have. 

b. If the number following the last place is 5 or larger, add 1. 

c. 

Drop all other numbers following the last decimal place.


Examples: 

1. Express these decimals to the nearest thousandth: 

(a) 0.6254 (6) 3.1416 ( 

Fig. 31. All dimensions are in inches. 


Find the ever-all length of the fitting in Fig. 31. 

Over-all length - 0.625 + 2.500 + 0.625 

0.625 

2.500 

0.625 

3 . 

750 

Over-all length = 3.750 in. 

Am. 

To add decimals, arrange the numbers so that all decimal 

points are on the same line. 

Examples: 

Add 

1. 4.75 + 3.25 + 6.50 

2. 3.055 + 0.257 + 0.125

Decimals in Aviation 23 

3. 18.200 + 12.033 + 1.800 + 7.605 

4. 0.003 + 0.139 + 0.450 + 0.755 

6. Find the over-all length of the fitting in Fig. 32. 

A =0.755" 

- ----- C 

- 

Fig. 32. 

C = 3.125" 

D= 0.500" 

E = 0.755" 

6. The thickness gage in Fig. 33 has six tempered-steel 

leaves of the following thicknesses: 

(b) 2 thousandths 

(a) l thousandths 

(c) 6 thousandths (/) 15 thousandths 

(r) 3 thousandths 

(d) 4 thousandths 

What is the total thickness of all six leaves? 

Fig. 33. Thickness gage. 

7. A thickness gage has H tempered 

steel leaves of the 

following thicknesses: 0.0015, 0.002, 0.003, 0.004, 0.006, 

0.008, 0.010, and 0.015. 

a. What is their total thickness? 

b. Which three leaves would add up to 

thousandths? 

c. Which three leaves will give a combined 

thickness of 10^ thousandths? 

B. Subtraction of Decimals. 

In Fig. 34, one dimension has been omitted. 

\V-I.I2S" J 

1.375" ->| 

Fig. 34.



Method: 

Find the missing dimension in Fig. 34. 

A = 1.375 - 1.125 

1 . 375 

-1.125 

0.250 in. Am. 

In subtracting decimals, make sure that the decimal points 

are aligned. 

Examples: 

Subtract 

1. 9.75 - 3.50 2. 2.500 - 0.035 

3. 16.275 - 14.520 4. 0.625 - 0.005 

5. 48.50 - 0.32 6. 1.512 - 0.375 

7. What are the missing dimensions in Fig. 35 ? 

Z/25"---4< A ^- 1.613"-+- 

6.312" 

Fig. 35. Front wing spar. 

Do these examples: 

8. 4.325 + 0.165 - 2.050 

9. 3.875 - 1.125 + 0.515 + 3.500 

10. 28.50 - 26.58 + 0.48 - 0.75 

11. 92.875 + 4.312 + 692.500 - 31.145 - 0.807 

12. 372.5 + 82.60 - 84.0 - 7.0 - 6.5 

Job 3: Multiplication 

of Decimals 

The multiplication of decimals is 

just as easy as the 

multiplication of whole numbers. Study the illustrative 

example carefully.



Find the total height of 12 sheets of aircraft sheet aluminum, 

B. and S. gage No. 20 (0.032 in.). 

I2$heefs ofB.aS #20 

Fi g . 36. 

Method: 

Multiply 0.032 by 12. 

0.032 in. 

_X12 

064 

J32 

0.384 in. Am. 

a. Multiply as usual. 

6. Count the number of decimal places in the numbers being 

multiplied. 

c. Count off the same number of decimal places in the answer, 

starting at the extreme right. 

Examples: 

Express all answers to the nearest hundredth : 

1. 0.35 X 2.3 2. 1.35 X 14.0 

3. 8.75 X 1.2 4. 5.875 X 0.25 

6. 3.1416 X 0.25 6. 3.1416 X 4 X 4 

7. A dural 1 sheet of a certain thickness weighs 0.174 Ib. 

per sq. ft. What is the weight of a sheet whose area is 

16.50 sq. ft.? 

8. The price per foot of a certain size of seamless steel 

tubing is $1.02. What is the cost of 145 ft. of this tubing? 

9. The Grumman G-21-A has a wing area of 375.0 

sq. ft. If each square foot of the wing can carry an average 

1 The word dural is a shortened form of duralumin and is commonly used in 

the aircraft trades.


weight of 21.3 lb., 

carry ? 

how many pounds can the whole plane 

Fis. 37. Grumman G-21-A, an amphibian monoplane. (Courtesy of Aviation.) 

Job 4: Division of Decimals 

A piece of flat stock exactly 74.325 in. long is to be sheared into 

15 equal parts. What is the length of each part to the nearest 

thousandth of an inch ? 

74.325" 

Fi 9 . 38. 


Divide 74.325 by 15. 

4.9550 

15)74.325^ 

60 

14~3 

13 5 

75 

75~ 

75 

Each piece will be 4.955 in. long. 

Ans.


Examples: 

Express all 

answers to the nearest thousandth: 

1. 9.283 -T- 6 2. 7.1462 ^9 3. 2G5.5 -r 18 

4. 40.03 -T- 22 5. 1.005 -5-7 6. 103.05 ~- 37 

Express answers to the nearest hundredth: 

7. 46.2 ~ 2.5 8. 7.36 -5- 0.8 9. 0.483 ~ 4.45 

10. 0.692 4- 0.35 11. 42 -r- 0.5 12. 125 -f- 3.14 

13. A f-in. rivet weighs 0.375 Ib. How many 

rivets are 

there in 50 Ib. ? 

14. Find the wall thickness 

/ of the tubes in Fig. 39. 

15. A strip of metal 16 in. 

long is to be cut into 5 equal 

(ct) 

parts. What is the length of 

(b) 

' 

9> 

each part to the nearest 

thousandth of an inch, allowing nothing for each cut of the 

shears ? 

Job 5: Changing Fractions to Decimals 

Any fraction can be changed into a decimal by dividing 

the numerator by the denominator. 


Change 

to a decimal. 

4 = ~ 

6 

0.8333-f An*. 

6)5.0000""" 

Hint: The number of decimal places in the answer depends on 

the number of zeros added after the decimal point. 

Change f to a decimal accurate to the nearest thousandth. 

0.4285+ = 0.429 Arts. 

f = 7)3.0000""


Examples: 

1. Change these fractions to decimals accurate to the 

nearest thousandth: 

() f (6) I (


Decimal Equivalents 

,015625 

.03125 

.046875 

.0625 

.078125 

.09373 

.109375 

-125 

.140625 

.15625 

.171875 

.1875 

.203125 

.21875 

.234375 

.25 

-.265625 

-.28125 

.296875 

-.3125 

-.328125 

.34375 

.359375 

.375 

.390625 

.40625 

-.421875 

r-4375 

K453I25 

.46875 

.484375 

5 

-.515626 

-.53125 

K540875 

-.5625 

K578I25 

-.59375 

K609375 

-.625 

K640625 

-.65625 

K67I875 

-.6875 

K 7031 25 

.71875 

.734375 

.75 

.765625 

.78125 

.796875 

.8125 

^ 

-.84375 

'28125 

.859375 

.875 

.890625 

.90625 

.921875 

.9375 

.953125 

.96875 

. 984375 

Fig. 41. 

important. See how quickly you can do the following 

examples. 

Examples: 

Change these fractions to decimals: 

i. f 

6. eV 

9. ifi 

2. 'i 

6. Jf 

10. ft 

3. i 

7. A 

11. h 

4- -& 

8. A 

12. e 

Change these fractions to decimals accurate to the 

nearest tenth: 

13. 14. 15. i% 16. 

Change these fractions to decimals accurate to the 

nearest hundredth: 

17. 18. M 19.fi 20.fl


Change these fractions to decimals accurate to the nearest 

thousandth : 

21. 

M. 

22. & 23. -& 24. 

Change these mixed numbers to decimals accurate to the 

nearest thousandth: 

Hint: Change the fraction only, not the whole number. 

25. 3H 26. 8H 27. 9& 28. 3ft 

Certain fractions are changed 

to decimals so often 

that it is worth remembering their decimal equivalents. 

Memorize the following fractions and their decimal 

equivalents to the nearest thousandth: 

i = 0.500 i = 0.250 f = 0.750 

i = 0.125 | = 0.375 f = 0.625 % = 0.875 

-iV = 0.063 jfe = 0.031 ^f = 0.016 

B. Changing Decimals to Ruler Fractions. The decimal 

equivalent chart can also be used to change any decimal 

to its nearest ruler fraction. This is 

extremely important 

in metal work and in the machine shop, as well as in many 

other jobs. 


Change 0.715 to the nearest ruler fraction. 

From the decimal equivalent chart we can see that 

If = .703125, f| - .71875 

0.715 lies between f and ff , 

but it is nearer to -f-g-. 

Ans. 

Examples: 

1. Change 

these decimals to the nearest ruler fraction: 

(a) 0.315 (b) 0.516 (c) 0.218 (rf) 0.716 (e) 0.832 

2. Change these decimals to the nearest ruler fraction: 

(a) 0.842 (6) 0.103 (c) 0.056 (d) 0.9032 (e) 0.621 

3. Change these to the nearest 64th : 

(a) 0.309 (b) 0.162 (c) 0.768 (d) 0.980 (e) 0.092


4. As a mechanic you are to work from the drawing in 

Fig. 42, but all you have is a steel rule graduated in 64ths. 

Convert all dimensions to fractions accurate to the nearest 

64th. 

-0< ^44- 

_2 

Fig. 42. 

5. Find the over-all dimensions in Fig. 42 (a) in decimals; 

(6) in fractions. Fig. 43. Airplane turnbuclde. 

6. Here is a table from an airplane supply catalogue 

giving the dimensions of aircraft turnbuckles. Notice how 

the letters L, A, D, ./, and G tell exactly what dimension is 

referred to. Convert all decimals to ruler fractions accurate 

to the nearest 64th.


7. A line 5 in. long is to be sheared into 3 equal parts. 

the length of each part to the nearest 64th of an 

What is 

inch ? 

Job 7: 

Tolerance and Limits 

A group of apprentice mechanics were given the job of 

cutting a round rod 2^ in. long. They had all worked from 

the drawing shown in Fig. 44. The inspector who checked 

their work found these measurements : 

(6) H (c) 2f (d) 

Should all 

pieces except e be thrown away? 

. 

2-'" >| Tolero,nce'/32 

Fig. 44. 

Round rod. 

Since it is 

impossible ever to get the exact size that a 

blueprint calls for, the mechanic should be given a certain 

permissible leeway. This leeway is called the tolerance. 

Definitions: 

Basic dimension is the exact size called for in a blueprint 

or working drawing. For example, 2-g- in. is the basic 

dimension in Fig. 44. 

Tolerance is the permissible variation from the basic 

dimension. 

Tolerances are always marked on blueprints. A tolerance 

of ^ means that the finished product will be acceptable 

even if it is as much as y 1^ in. greater or \- in. less than the 

basic dimension. A tolerance of 0.001 means that permissible 

variations of more and less than the basic size are 

acceptable providing they 

fall with 0.001 of the basic 

dimension. A tolerance of 

A'QA.I 

means that the finished 

part will be acceptable even if it is as much as 0.003 greater


than the basic dimension; however, it may only be 0.001 

less. 

Questions: 

1. What does a tolerance of -gV mean? 

2. What do these tolerances mean? 

(a) 0.002 (6) 0.015 , . 

+0.005 

(C) 

, , 

I* 

+0.0005 

, 

+0.002 

-0.001 

W (e) 

-0.0010 

-0.000 

3. What is meant by 

a basic dimension of 3.450 in. ? 

In checking the round rods referred to in Fig. 44, the 

inspector can determine the dimensions of acceptable pieces 

of work by adding the plus tolerance to the basic dimension 

and by subtracting the minus tolerance from the basic 

dimension. This would give him an upper limit and a lower 

limit as shown in Fig. 44a. Therefore, pieces measuring less 

than 2^| in. 

are not acceptable; neither are pieces measur- 

+Basf'c size = Z 1 //- 

-Upper limii-:2^2 =2j2 

>\ 

Fig. 44a. 

ing more than 2^-J in. As a result pieces a, 6, d, and e are 

passed, and c. is rejected. 

There is another way of settling the inspector's problem. 

All pieces varying from the basic dimension by more than 

3V in. will be rejected. Using this standard we find that 

piece d 

pieces a and 6 vary by only -fa; piece c varies by -g-; 

varies by 3^; piece e varies not at all. All pieces except 

c are therefore acceptable. The inspector knew that the 

tolerance was -&$ in. because it was printed on the 

drawing.

34 Mathematics for the >Av/at/on Tracfes 

Examples: 

1. The basic dimension of a piece of work is 3 in. and 

the tolerance is ^ in. Which of these pieces 

acceptable ? 

are not 

(a) %V (b) ff (c) 2-J ((/) Si (e) Sg^s- 

2. A blueprint gives a basic dimension of 2| in. arid 

tolerance of &$ in. Which of these pieces should be 

rejected? 

(a) 2|i (b) 2-$| (c) 2 :Vf (d) 2.718 (e) 2.645 

3. What are the upper and lower limits of a job whose 

basic dimension is 4 in., if the tolerance is 0.003 in.? 

4. What are the limits of a job where the tolerance is 

n nm 

; 

tf ê basic dimension is 2.375? 

U.Uul 

6. What are the limits on the length and width of the 

job in Fig. 

Fis. 44b. 


1. Express answers to the nearest hundredth: 

(a) 3.1416 X 2.5 X 2.5 (6) 4.75 - 0.7854 

(c) 4.7625 + 0.325 + 42 - 20.635 - 0.0072 

2. Convert these gages to fractions, accurate to the 

nearest 64th:


3, Often the relation between the parts of a fastening 

is 

given in terms of one item. For example, 

in the rivet in 

Fis. 45. 

Fig. 45, all parts depend on the diameter of the shank, as 

follows: 

R = 0.885 X A 

C = 0.75 X A 

B = 1.75 X A


Complete the following table: 

4. A 20-ft. length of tubing is to be cut into 7|-in. 

lengths. Allowing jV in. 

for each cut, how many pieces of 

tubing would result? What would be the length of the last 

piece ? 

5. Measure each of the lines in Fig. 45a to the nearest 

64th. Divide each line into the number of equal parts 

indicated. What is the length of each part as a ruler 

fraction ? 

(a) 

H 3 Equal par is 

H 5 Equal parts 

(c) h -I 6 Equal parts 

(d) 

Fig. 45a. 

4 Equal parts

Chapter III 

MEASURING LENGTH 

The work in the preceding chapter dealt with measuring 

lengths with the steel rule or the micrometer. The answers 

to the Examples have been given as fractions or as decimal 

parts of an inch or inches. However, there are many other 

units of length. 

Job 1 : Units of Length 

Would it be reasonable to measure the distance from New 

York to Chicago in inches? in feet? in yards? What unit is 

generally used? If we had only one unit of length, could it 

be used very conveniently for all kinds of jobs? 

In his work, a mechanic will frequently meet measurements 

in various units of length. Memorize Table 1. 

Examples: 

S^ft.? 

TABLE 1. 

LENGTH 

12 inches (in. or ") 

= 1 foot (ft. or ') 

8 feet = 1 yard (yd.) 

5,280 feet = 1 mile (mi.) 

1 meter = 89 inches (approx.) 

1. How many inches are there in 5 ft.? in 1 yd.? in 

2. How many feet are there in 3^ yd.? 

Similes? 

3. How many yards are there in 1 mile? 

4. How many inches are there in yV mile? 

in 48 in.? in 

6. Round rod of a certain diameter can be purchased 

at $.38 per foot of length. What is the cost of 150 in. of this 

rod? 

37


6. Change 6 in. to feet. 

Hint: Divide 6 by 12 and express in simplest terms. 

the answer as a fraction 

7. Change 3 in. to feet. Express the answer as a decimal. 

8. Change these dimensions to feet. Express the answers 

as fractions. 

(a) 1 in. (b) 2 in. (r) 4 in. (d) 5 in. 

9. Change these dimensions to decimal parts of a foot, 

accurate to the nearest tenth. 

(a) 6 in. (6) 7 in. (c) 8 in. (d) 9 in. 

0) 10 in. (/) 11 in. (g) 12 in. (//) 

13 in. 

10. Change the dimensions in Fig. 46 to feet, expressing 

the answers as decimals accurate to the nearest tenth of a 

foot. 

Fig. 46. 

11. What are the span and length of the Fairchild F-45 

(Fig. 47) 

in inches? 

< 

--30-5" 

-: 

s. 47. Fairchild F-45. (Courtesy of Aviation.)

Measuring Length 39 

Job 2: Perimeter 

r 

\ 

Perimeter simply means the distance around as 

shown 

Fig. 48. 

in Fig. 48. To find the perimeter of a figure of any number 

of sides, add the length 

of all 

the sides. 


Find the perimeter of the triangle in Fig. 49. 

Fig. 49. 

Perimeter = 2 + IT + 2^ in. 

Perimeter = 5f in. Anx. 

Examples: 

1. Find the perimeter of a triangle whose sides are 

3-g- in., (>rg in., 2j in. 

2. Find the perimeter of each of the figures in Fig. 50. 

All dimensions are in inches. 

(a) 

Fig. 50. 

(b)

40 Mat/iematics for the Aviation Trades 

3. Find the perimeter of the figure in Fig. 

accurately to the nearest 32nd. 

51. Measure 

Fis. 51. 

4. A regular hexagon (six-sided figure in which all sides 

are of equal length) measures 8^ 

perimeter in inches? in feet? 

Job 3: Nonruler Fractions 

in. on a side. What is its 

It should be noticed that heretofore we have added fractions 

whose denominators were always 2, 4, 6, H, 16, 32, or 

64. These are the denominators of the mechanic's most 

useful fractions. Since they are found on the rule, these 

fractions have been called ruler fractions. There are, however, 

many occasions where it is useful to be able to add or 

subtract nonruler fractions, fractions that are not found 

on the ruler. 


Find the perimeter of the triangle in Fig. 52. 

Perimeter = 

Sum = 15U ft. 

Ans.


Notice that the method used in the addition or subtraction 

of these fractions is identical to the method already 

learned for the addition of ruler fractions. It is sometimes 

harder, however, to find the denominator of the equivalent 

fractions. This denominator is called the least common 

denominator. 

Definition: 

The least common denominator (L.C.D.) of a group of 

fractions is the smallest number that can be divided exactly 

by each of the denominators of all the fractions. 

For instance, 10 is the L.C.D. of fractions -^ and \ because 

10 can be divided exactly both by 2 and by 5. Similarly 15 

is the L.C.D. of f and \. Why? 

There are various methods of finding the L.C.D. The 

or trial and error. What is the 

easiest one is by inspection 

L.C.D. of -gand 

^? Since 6 can be divided exactly by both 

2 and 3, (> is the L.C.D. 

Examples: 

1. Find the L.C.D. (a) Of i and i 

(6) Of i , f 

(

; O 


5. Find the total length in feet of the form in Fig. 53. 

6. Find the total length in feet of 3 boards which are 

ft., 8f ft., and 12f ft. long. 

7. Find the perimeter of the figure in Fig. 54. 

-- 

1 

= 

b l/l2 , 

f W c = 

_ Ct = / 3 /4 

Fi g . 54. 

8. Find the perimeter of the plate in Fig. 55. Express 

the answer in feet accurate to the nearest hundredth of 

a foot. 

9. The perimeter of a triangle 

is 12y^ ft. If the first side 

is 4-g- ft. and the second side is 2f ft., what is the length of 

the third side? 

10. Find the total length in feet of a fence needed to 

enclose the plot of ground shown in Fig. 56. 

Pis. 56.


Job 4: 

The Circumference of a Circle 

Circumference is a special word which means the distance 

around or the perimeter of a circle. There is 

absolutely no 

reason why the word perimeter could not be used, but it 

never is. 

A Few Facts about the Circle 

1. Any line from the center to the circumference is called 

a radiux. 

2. Any line drawn through the center and meeting the 

circumference at each end is called 

a diameter. 

3. The diameter is twice as long 

as the radius. 

4. All radii of the same circle are 

equal; all diameters of the same 

circle are equal. 

Finding the circumference of a 

circle is a little harder than finding: _ "V" V*"7 

Fig. 57. Circle, 

the distance around figures with 

straight sides. The following formula is used: 

Formula: C = 3.14 X D 

where C = circumference. 

D = diameter. 

The "key number" 3.14 is 

used in finding the circumference 

of circles. No matter what the diameter of the 

circle is, to find its circumference, multiply the diameter by 

the "key number/' 3.14. This is 

only an approximation of 

the exact number 3.1415926+ which has a special name, TT 

(pronounced pie). Instead of writing the long number 

3. 1415926 +, it is easier to write TT. The circumference of a 

circle can therefore be written 

C = 

X D


If a greater degree of accuracy is required, 3.1416 can be 

used instead of 3.14 in the formula. The mechanic should 

practically never have any need to go beyond this. 


Find the circumference of a circle whose diameter is 

3.5 in. 

Fig. 58. 

Given: D = 3.5 in. 

Find : Circumference 

Examples: 

4 in. 

C = 3.14 X D 

C = 3.14 X 3.5 

C = 10.99 in. Ans. 

1. Find the circumference of a circle whose diameter is 

2. What is the distance around a pipe whose outside 

diameter is 2 in. ? 

3. A circular tank has a diameter of 5 ft. What is its 

circumference ? 

4. Measure the diameter of the circles in Fig. 

nearest 32nd, and find the circumference of each. 

59 to the 

(C)


5. Estimate the circumference of the circle in Fig. 60a. 

Calculate the exact length after measuring the diameter. 

How close was your estimate? 

6. Find the circumference of a circle whose radius is 3 in. 

Hint: First find the diameter. 

7. What is the total length in feet of 3 steel bands which 

must be butt-welded around the barrel, as shown in Fig. 60& ? 

Fi9. 60a. Fig. 60b. 

8. What is the circumference in feet of a steel plate whose 

radius is 15-g- in.? 

9. What is the circumference of a round disk whose 

diameter is 1.5000 in.? Use TT = 3.1416 and express the 

answer to the nearest thousandth. 


1. The Monocoupe shown in Fig. 61 has a length of 

246.5 in. What is its length in feet? 

Fig. 61 . 

Monocoupe high-wing monoplane. (Courtesy of Aviation.)


2. Complete 

this table: 

3. Find the missing dimensions in Fig. 62. 

Perimeter = 18i% in. 

a = b 

c = d = ? 

= 4M in. 

4. Find the inner and outer circumferences of the circular 

Fig. 63. 

disk shown in Fig. 03. Express the answers in decimals 

accurate to the nearest thousandth. 

5. Find the perimeter of the flat plate shown in Fig. 64.

Copter IV 

THE AREA OF SIMPLE FIGURES 

The length of any object can be measured with a rule 

It is impossible, however, to measure area so directly and 

simply as that. In the following pages, you will meet 

geometrical shapes like those in Fig. 65. 

Circle Rectangle Square 

Triangle 

Fi g . 65. 

Trapezoi'd 

Each of these shapes will require a separate formula and 

some arithmetic before its area can be found. You should 

know these formulas as well as you know how to use a rule. 

A mechanic should also know that these are the crosssectional 

shapes of most common objects, such as nails, 

beams, rivets, sheet metal, etc. 

Job 1 : 

Units of Area 

Would you measure the area of a small piece of metal in 

square miles? Would you measure the area of a field in 

square inches? The unit used in measuring area depends 

on the kind of work being done. Memorize this table: 

47


TABLE 2. 

AREA 

144 square inches = 1 square foot (sq. ft.) 

9 square feet = 1 square yard (sq. yd.) 

640 acres = 1 square mile (sq. mi.) 

4,840 square yards = 1 acre 

i 

n 

[ /' J U /'- J 

Examples: 

Fig. 66. 

"" 

1. How many square inches are there in 3 sq. ft.? 1 sq. 

yd.?isq. ft.?2|-sq. yd.? 

2. How many square feet are there in 4 sq. yd.? 1 sq. 

mile? 1 acre? 

3. How many square yards are there in 5 square miles? 

l,000sq. ft.? 60 acres? 

4. If land is 

bought at $45.00 an acre, what is the price 

per square mile? 

5. What decimal part of a square foot is 72 sq. in. ? 

36 sq. in.? 54 sq. in.? 

Job 2: 

The Rectangle 

A. Area. The rectangle 

is the cross-sectional shape of 

many beams, fittings, and other common objects. 

"T 

I 

Length 

Fig. 66a.--Rectangle.

77e Area of Simple Figures 49 

A Few Facts about the 

1. 

Opposite sides are equal 

2. All four angles are right angles. 

3. The sum of the angles is 360. 

Rectangle 

to each other. 

4. The line joining opposite corners is called a diagonal. 

Formula: A ^ L X W 

where A = area of a rectangle. 

L = length. 

W = width. 


Find the area of a rectangle whose length 

is 14 in. and whose 

width is 

3 in. 

Given: L = 14 in. 

W = 3 in. 

Find: Area 

Examples: 

A = L X W 

A = 14 X 3 

A = 42 sq. in. 

Ans. 

Find the area of these rectangles: 

1. L = 45 in., W = 16.5 in. 

2. L = 25 in., W = 5^ in. 

3. IF = 3.75 in., L = 4.25 in. 

4. L = sf ft., rr = 2| ft. 

5. Z = 15 in., W = 3| in. 

6. 7, = 43 ft., TF - s ft. 

7. By using a rule, find the length and width (to the 

(b) 

(ct) (c) (d) 

Fig. 67.


nearest 16th) of the rectangles in Fig. 67. Then calculate 

the area of each. 

8. Find the area in square feet of the airplane wing shown 

in Fig/ 68. 

Trailin 

A/Te ran 

\ I Aileron 

V* 

Fig. 68. 

'- 

Leading edge 

Airplane wing, top view. 

9. Calculate the area and perimeter of the plate shown 

in Fig. 69. 

i 

Fig. 69. 

B. Length and Width. To find the length or the width, 

use one of the following formulas: 

Formulas: L 

w 

A 

L 

where L = length. 

A area. 

W = width. 


The area of a rectangular piece of sheet metal is 20 sq. ft.; 

its width is ^ ft. What is its length? 

Given: A = 20 sq. ft. 

W = 2i ft.

The Area of Simple Figures 51 

Find: 

Length 

L = y IF 

Check: yt=L 

20 

2i 

= 20 X | 

= Hft. Ann. 

H' = 8X 2 

- 20 sq. ft. 

Examples: 

1. The area of a rectangular floor is 75 sq. ft. What is the 

length of the floor if its width is 7 ft. in. ? 

2-5. Complete this table by finding the missing dimension 

of these rectangles: 

6. What must be the width of a rectangular beam whose 

cross-sectional area is 10.375 sq. in., and whose length is 

5, 3 in. as shown in Fig. 70? 

7. The length (span) of a rectangular wing is 17 ft. 6 in.; 

ft. What is the width 

its area, including ailerons, is 50 sq. 

(chord) of the wing?


Job 3: Mathematical Shorthand: Squaring a Number 

A long time ago you learned some mathematical shorthand, 

-for instance, + (read "plus") is shorthand for add; 

(read "minus") is shorthand for subtract. Another 

important shorthand symbol in mathematics is the small 

2 

number two ( ) 

written near the top of a number: 5 2 (read 

"5 squared") means 5 X 5; 7 2 (read "7 squared") means 

7X7. 

You will find this shorthand very valuable in working 

with the areas of circles, squares, and other geometrical 

figures. 


What is 7 squared? 

7 squared = 7 = 2 7 X 7 = 49 Ans. 

Examples: 

Calculate: 

What is () 9 2 ? (6) (3.5) 2 ? (c) (i) 2 ? 

(a) 9 2 = 9 X 9 = 81 Ans. 

(6) (3.5) 2 = 3.5 X 3.5 = 12.25 Ans. 

to (f) 2 = tXf = J = 2i Ans. 

1. 5 2 2. 3 2 3. (2.5) 2 4. I 2 

5. (9.5) 2 6. (0.23) 2 7. (2.8) 2 8. (4.0) 2 

Reduce answers to lowest terms whenever necessary: 

9. (|) 2 10. (IY 11. (f) 

2 

13. (|) 

2 

Calculate: 

14. (f) 

2 

12. (A) 2 

15. ( 

l^) 2 16. (V-) 2 

17. (2*)' 18. (3i) 2 19. (4^) 2 20. 

21. (9f) 2 22. (12^)* 

Complete: 

23. 3 X (5 

s 

) 

= 24. 4 X (7 

2 

) 

= 

2 

26. 0.78 X (6 

= 2 

) 26. 4.2 X ( 

27. 3.1 X (2 = 2 ) 

28. 3.14 X (7 = 2 ) 

) 

=

The Area of Simple Figures 

B 

53 

Fis. 71. 

29. Measure to the nearest 64th the lines in Fig. 

complete this table: 

71. Then 

AC = 

BC = 

AB = 

Is this true: AC 2 + BC 2 = AB 2 ? 

30. Measure, to the nearest (>4th, the lines in Fig. 72. 

Then complete this table: 

R g 

. 72. 

AC - 

no, = 

AH = 

Is this true: AC* + BC 2 = AB 2 ? 

Is this true: stagger 2 + gap 2 = strut 

2 ?


Job 4: Introduction to Square Root 

The following squares were learned from the last job. 

TABLE 8 

I 2 = 1 (i 2 = 86 

& = 4 7 

2 - 49 

3 2 - 8 2 - 64 

42 = 1(5 92 = 81 

52 = 25 10 

2 - 100 

Find the answer to this question in Table 3: 

What number when multiplied by itself equals 49? 

The answer is 7, which is said to be the square root of 41), 

written \/49. The mathematical shorthand in this case is 

entire question can be 

V (read "the square root of ") 

. The 

written 

What is V49? The answer is 7. 

Check: 7X7= 49. 

Examples: 

1. What is the number which when multiplied by itself 

equals 64? This answer is 8. Why? 

* 

2. What number multiplied by itself equals 2.5 ? 

3. What is the square root of 100? 

4. What is V36? 

5. Find 

(a) V_ (6) VsT (r) VlO_ 

(


Job 5: 

The Square Root of a Whole Number 

So far the square roots of a few simple numbers have been 

found. There is, however, a definite method of finding the 

square root of any whole number. 


What is the square root of 1,156? 

3 Am. 

9 

64) 256 

256 

Method: 

Check: 34 X 34 = 1,156 

a. Separate the number into pairs starting 

from the right: 

A/11 56 

b. \/H 

lies between 3 and 4. Write the 

smaller number 3, above the 11: 

c. Write 3 2 or 9 below the 11: 

d. Subtract and bring down the next pair, 56 

e. Double the answer so far obtained 

(3 X * = 6). 

3 

56 

\Xli~56 

9 

_ 

A/11 56 

9 

__ 

2 56 

3_ 

\XlF56 

Write 6 as shown : 

/. Using the 6 just obtained as a trial divisor, 

divide it into the 25. Write the answer, 4, 

as shown: 

3 

__ 

A/11 56 

9

! 

the 

, 15. 


g. Multiply the 64 by the 4 just obtained 

and write the product, 256, as shown: 

3 4 Ans. 

56 

9 

64) 2 56 

2 56 

h. Since there is no remainder, the square root of 1,156 is 

exactly 34. 

Check: 34 X 34 - 1,156. 

Examples: 

1. Find the exact square root of 2,025. 

Find the exact square root of 

2. 4,225 3. 1,089 4. 625 5. 5,184 

What is 

the exact answer? 

6. V529 7. V367 8. \/8,4(>4 9. Vl~849 

10. Find the approximate square root of 1,240. Check 

your answer. 

Hint: Work as explained and ignore the remainder. To 

check, square your answer and add 

connection is 

length 

I 

remainder. 

Find the approximate square root 

of 

5 11. 4,372 12. 9,164 

13. 3,092 14. 4,708 

9,001 16. 1,050 

17. 300 18. 8,000 

Fi9 ' 73 ' 19. Study Fig-. 73 carefully. What 

of its sides? 

there between the area of this square and the 

Job 6: 

The Square Root of Decimals 

Finding the square root of a decimal is very much like 

finding the square root of a whole number. Here are two 

rules:


Rule 1. The grouping of numbers into pairs should 

always be started from the decimal point. For instance, 

Notice that a zero is 

362.53 is 

paired as 3 62. 53 

71.3783 is 

paired as 71. 37 83 

893.4 is paired as 8 93. 40 

15.5 is paired as 15. 50 

added to complete any incomplete, 

pair on the right-hand side of the decimal point. 

Rule 2, The decimal point of the answer is directly above 

the decimal point of the original number. 

Two examples are given below. Study them carefully. 


Find V83.72 

9.1 Ans. 

-V/83.72 

81 

181) 2 72 

1JU " 

91 

Check: 9.1 X 9.1 = 82.81 

Remainder = +.91 

83 . 72 

Find the square root of 7.453 

2.7 3 Ans. 

V7.45 30 

J 

47)^45 

329 

543) 16 30 

16 29 _ 

Check: 2.73 X 2.73 = 7.4529 

Remainder = +.0001 

7.4530


Examples: 

1. What is the square root of 34.92? Check your 

answer. 

What is 

the square root of 

* 

2. 15.32 

3. 80.39 

4. 75.03 5. 342.35 

What is 

7. \/4 1.35 

10. VT91.40 11. A/137.1 12. 

7720 9. V3.452 

27.00 13. V3.000 

Find the square 

14. 462.0000 

16. 4.830 

root to the nearest tenth: 

15. 39.7000 

17. 193.2 

18. Find the square root of jj 

to the nearest tenth. 

Hint: Change y to a decimal and find the square root of 

the decimal. 

Find the square root of these fractions to the nearest 

hundredth: 

19. 20. 22. i 23. 1 

24. Find the square 

75.00 

root of 

.78 

to the nearest tenth. 

Job 7: 

The Square 

A. The Area of a Square. The square is really a special 

kind of rectangle where all sides are equal in length.


A Few Facts about the Square 

1. All four sides have the same length. 

2. All four angles are right angles. 

3. The sum of the angles 

is 360. 

4. A line joining two opposite corners is called a diagonal. 

Formula: A - S 

2 

= S X S 

where N means the side of the square. 

Examples: 


Find the area of a square whose side is 

Given : *S = 5^ in. 

Find: Area 

A = 2 /S 

A = (5i)* 

A = V X V- 

A = -HP 30-i sq. in. Ans. 

Find the area of these squares: 

5^ in. 

1. side = 2i in. 2. side = 5i ft. 3. side = 3.25 in. 

4-6. Measure the length of the sides of the squares 

shown in Fig. 75 to the nearest 32nd, and find the area 

of each: 

Ex. 4 Ex. 5 Ex. 6 

Fi 9 . 75. 

7-8. Find the surface area of the cap-strip gages shown 

in Fig. 76.

. 

2 


Efe 

ST~ 

L /" r- 

--~,| |< 

^--4 

Ex. 7 Ex. 8 

Fig. 76. Cap-strip sages. 

_t 

9. A square piece of sheet metal measures 4 ft. 6 in. on 

a side. Find the surface area in (a) square inches; (6) square 

feet. 

10. A family decides to buy linoleum at $.55 a square 

yard. What would it cost to cover a square floor measuring 

12 ft. on a side? 

B. The Side of a Square. To find the side of a square, use 

the following formula. 

Formula: S = \/A 

where 8 = side. 

A = area of the square. 


A mechanic has been told that he needs a square beam whose 

cross-sectional area is 6. 

beam? 

Given: A = 6.25 sq. in. 

Find: 

Side 

5 sq. in. What are the dimensions of this 

s = 

8 = \/25 

S = 2.5 in. 

Ans. 

Check: A = 8 

2 = 2.5 X 2.5 = 6.25 sq. in.


Method: 

Find the square 

root of the area. 

Examples: 

Find to the nearest tenth, the side of a square whose area is 

1. 47.50 sq. in. 

3. 8.750 sq. in. 

2. 24.80 sq. ft. 

4. 34.750 sq. yd. 

5-8. Complete the following table by finding 

the sides in 

both feet and inches of the squares whose areas are given: 

Job 8: 

The Circle 

A. The Area of a Circle. The circle is the cross-sectional 

shape of wires, round rods, bolts, rivets, etc. 

Fig. 77. Circle. 

where A 

/) 2 

D 

Formula: A = 0.7854 X D 2 

area of a circle. 

D XD. 

diameter.



Find the area of a circle whose diameter is 

Given: D = 3 in. 

Find: A 

A = 0.7854 X D 2 

A = 0.7854 X 3 X 3 

A = 0.7854 X 9 

^4 = 7.0686 sq. in. Ans. 

3 in. 

Examples: 

Find the area of the circle whose diameter is 

1. 4 in. 2. i ft. 3. 5 in. 

4. S ft. 5. li yd. 6. 2| mi. 

7-11. Measure the diameters of the circles shown in 

Fig. 78 to the nearest Kith. Calculate the area of each 

circle. 

Ex.7 Ex.8 Ex.9 Ex.10 Ex.11 

12. What is the area of the top of a piston whose diameter 

is 4 -- in. ? 

13. What is the cross-sectional area of a |-in. aluminum 

rivet ? 

14. Find the area in square inches of a circle whose 

radius is 

1 ft. 

16. A circular plate has a radius of 2 ft. (> in. Find (a) the 

area in square feet, (b) the circumference in inches. 

B. Diameter and Radius. The diameter of a circle can 

be found if the area is known, by using this formula:


where D = diameter. 

A = area. 

Formula: D 


Find the diameter of a round bar whose cross-sectional area is 

3.750 sq. in. 

Given: A = 3.750 sq. 

Find : Diameter 

in. 

- 

\ 0.7854 

/) 

- /A0 " 

\ 0.7854 

J) = V4.7746 

D = 

Check: A = 0.7854 X D 2 = 0.7854 X 2.18 X 2.18 = 3.73 + 

sq. in. 

Why doesn't the answer check perfectly? 

Method: 

a. Divide the area by 0.7S54. 

6. Find the square root of the result. 

Examples: 

1. Find the diameter of a circle whose area is 78.54 ft. 

2. What is the radius of a circle whose area is 45.00 

sq.ft.? 

. 

3. The area of a piston 

is 4.625 sq. in. What is its 

diameter ? , 

*/ 

sectional area of 0.263 sq. in. () y 

What is the diameter of the wire? 

4. A copper wire has a crossf 

HJ 

Section A -A 

(6) What is its radius? Are*1.02Ssq.in. 

6. A steel rivet has a cross-sec- 

tional area of 1.025 sq. in. What is its diameter (see Fig. 71)) ? 

' 

9 *


6-9. Complete 

this table: 

10. Find the area of one side of the washers shown in 

Fig. 80. Fig. 80. 

Job 9: 

The Triangle 

So far we have studied the rectangle, the square, and the 

circle. The triangle is another simple geometric figure often 

met on the job. 

A Few Facts about the Triangle 

1. A triangle has only three sides. 

2. The sum of the angles of a triangle 

is 180. 

Fig. 81. 

Base 

Triangle.

Tfie Area of Simple Figures 65 

3. A triangle having one right angle is called a right 

triangle. 

4. A triangle having all sides of the same length is called 

an equilateral triangle. 

5. A triangle having two equal sides is called an isosceles 

triangle. 

Right Equilaferal Isosceles 

Fig. 81 a. Types of triangles. 

The area of any triangle can be found by using this 

formula: 

where b 

a 

= the base. 

= the altitude. 

Formula: A = 

l/2 

X b X a 


Find the area of a triangle whose base is 7 in. long and whose 

altitude is 3 in. 

Given: 6 = 7 

Find: 

Examples: 

a = 3 

Area 

A = | X b X a 

A = 

i- X 7 X 3 

A = -TT 

= 10-g- sq. in. Ans. 

1. Find the area of a triangle whose base is 8 in. and 

whose altitude is 5 in. 

2. A triangular piece of sheet metal has a base of 16 in. 

and a height of 5^ in. What is its area? 

3. What is the area of a triangle whose base is 8-5- ft. and 

whose altitude is 2 ft. 

3 in.?


4-6. Find the area of the following triangles 

: 

7-9. Measure the base and the altitude of each triangle 

in Fig. 82 to the nearest 64th. Calculate the area of each. 

Ex.7 

Ex.8 

Fig. 82. 

Ex.9 

10. Measure to the nearest 04th and calculate the area 

of the triangle in Fig. 83 in the three different ways shown 

c 

Fig. 83. 

in the table. Does it make any difference which side is 

called the base? 

Job 10: The Trapezoid 

The trapezoid often appears as the shape of various parts 

of sheet-metal jobs, as the top view of an airplane wing, as 

the cross section of spars, and in many other connections.

Tfie Area of Simple Figures 67 

A Few Facts about the Trapezoid 

1. A trapezoid has four sides. 

2. Only one pair of opposite sides is parallel. These sides 

are called the bases. 

|< 

Base (bj) * 

k Base (b 2 ) ->| 

Fig. 84. Trapezoid. 

3. The perpendicular distance between the bases is 

called the altitude. 

Notice how closely the formula for the area of a trapezoid 

resembles one of the other formulas already studied. 

Formula: A - 

l/2 

X a X (b t + b 2 ) 

where a = the altitude. 

61 

= one base. 

6 2 

= the other base. 


Find the area of a trapezoid whose altitude is 

bases are 9 in. and 7 in. 

Given: a = 6 in. 

= 61 7 in. 

6 2 

= 9 in. 

Find: Area 

6 in. and whose 

Examples: 

A = i X a X (fci + 6 2 ) 

A = 1 X 6 X (7 + 9) 

A = i x 6 X 16 

A 48 sq. in. Atts. 

1. Find the area of a trapezoid whose altitude is 10 in, 

and whose bases are 15 in. and 12 in.


2. Find the area of a trapezoid whose parallel sides are 

1 ft. 3 in. and 2 ft. 6 in. and whose altitude is J) in. Express 

your answer in (a) square feet (&) square inches. 

3. Find the area of the figure in Fig. S4a, after making all 

necessary measurements with a rule graduated in 3nds. 

Estimate the area first. 

Fig. 84a. 

4. Find the area in square feet of the airplane wing, 

including the ailerons, shown in Fig. Mb. 

^^ 

Leading edge 

/5 -'6"- 

>| 

/ A Heron Aileron 

IQ'-9'L 

Trailing edge 

Fi s . 84b. 

5. Find the area of the figures in Fig. 84c. 

. / 

r*' 

(a) 

Fis. 84c. 

Job 11 : 

Review Test 

1. Measure to the nearest 32nd all dimensions indicated 

by letters in Fig. 85.

The, Area of Simple Figures 69 

4-*- E H* F *K

Gapter V 

VOLUME AND WEIGHT 

A solid is the technical term for anything that occupies 

space. For example, a penny, a hammer, and a steel rule 

are all solids because they occupy a definite space. Volume 

is 

the amount of space occupied by any object. 

Job 1 : 

Units of Volume 

It is too bad that there is no single unit for measuring 

all kinds of volume. The volume of liquids such as gasoline 

is generally measured in gallons; the contents of a box is 

measured in cubic inches or cubic feet. In most foreign 

countries, the liter, which is about 1 quart, 

is used as the unit 

of volume. 

However, all units of volume are interchangeable, and 

any one of them can be used in place of any other. Memorize 

the following table: 

TABLE 4. 

VOLUME 

1,728 cubic inches = 1 cubic foot (cu. ft.) 

27 cubic feet 1 cubic yard (cu. yd.) 

2 pints 

= 1 quart (qt.) 

4 quarts 

= 1 gallon (gal.) 

281 cubic inches = 1 

gallon (approx.) 

1 cubic foot = 1\ gallons (approx.) 

1 liter = 1 auart faoorox.)

Volume and Weight 71 

Fig. 87. 

Examples: 

1. How many cubic Inches are there in 5 cu. ft.? in 

1 cu. yd.? in ^ cu. ft.? in 3-j- cu. yd.? 

2. How many pints are there in (> 

qt.? in 15 gal.? 

3. How many cubic inches are there in C 2 qt. ? in ^ gal. ? 

4. How many gallons are there in 15 cu. ft. ? in 1 cu. ft. ? 

Job 2: The Formula for Volume 

88 below shows three of the most common 

Figure 

geometrical solids, as well as the shape of the base of each. 

Solid 

~ TTfjl 

h !'! 

Cylinder Box Cube 

Boise 

Circfe 

Rectangle Square 

Fi 3 . 88. 

The same formula can be used to find the volume of a 

cylinder, a rectangular box, or a cube.


Formula: V = A X h 

where V = volume. 

A = area of the base. 

h = height. 

Notice that it will be necessary to remember the formulas 

for the area of plane figures, in order to be able to find the 

volume of solids. 


Find the volume in cubic inches of a rectangular box whose 

base is 4 by 7 in. and whose height 

is 9 in. 

Given : 

Base: rectangle, L = 7 in. 

Find: 

a. Area of base 

6. Volume 

H' = 4 in. 

h = 9 in. 

a. Area = L X W 

Area ^7X4 

Area = 28 sq. in. 

b. Volume = A X h 

Volume = 28 X 9 

Volume = 252 cu. in. 

Arts. 

Examples: 

1. Find the volume in cubic inches of a box whose 

height is 15 in. arid whose base is 3 by 4-^ in. 

3^ in. 

2. Find the volume of a cube whose side measures 

3. A cylinder has a base whose diameter is 2 in. Find 

its volume, if it is 3.25 in. high. 

4. What is the volume of a cylindrical oil tank whose 

base has a diameter of 15 in. and whose height 

is 2 ft.? 

Express the answer in gallons. 

6. How many cubic feet of air does a room 12 ft. 6 in. 

by 15 ft. by 10 ft. 3 in. contain?


6. Approximately how many cubic feet of baggage can 

be stored in the plane wing compartment shown in Fig. 89? 

7. How many gallons of oil can be contained in a rectangular 

tank 3 by 3 by 5 ft. ? 

Hint: Change cubic feet to gallons. 

8. What is the cost, at $.19 per gal., of enough gasoline 

to fill a circular tank the diameter of whose base is 8 in. 

and whose height 

is 15 in.? 

9. How many quarts of oil can be stored in a circular 

tank 12 ft. 3 in. long 

3 ft. 6 in. ? 

if the diameter of the circular end is 

10. An airplane has 2 gasoline tanks, each with the 

specifications shown in Fig. !)0. How many gallons of fuel 

can this plane hold? 

[

. 1342 

711 

542 

450 

. . 


TABLE 5. COMMON WEIGHTS 

50 

. 

. 

. 

. .. 02.5 

Metals, Ib. per cu. ft. 

Woods,* Ib. per cu. ft. 

Aluminum 162 Ash 

physical properties affect the weight. 

Copper Mahogany 

53 

Dural 175 Maple 

49 

Iron (cast) Oak 

52 

Lead . Pine 

45 

Platinum. . 

Spruce 27 

Steel 490 

Air 

0.081 

Water.... 

* The figures for woods are approximate, since variations


b. Weight = V X unit weight 

Weight = 44- X 52 

Weight - 234 11). A ?iff. 

Notice that the volume is calculated in the same units 

(cubic feet) as the table of unit weights. Why 

is this 

essential ? 

Examples: 

1. Draw up a table of weights per 

materials given 

examples. 

Find the weight 

of 

in Table 5. Use 

cubic inch for all the 

this table in the following 

of each of these materials: 

2. 1 round aluminum rod 12 ft. 

long and with a diameter 

in. 

3. 5 square aluminum rods, l| by 1-J- in., in 12-ft. 

lengths. 

4. 100 square hard-drawn copper rods in 12-ft. lengths 

each J by J in- 

5. 75 steel strips each 4 by f in. in 25-ft. lengths. 

Find the weight of 

18 ft. 

6. A spruce beam 1^ by 3 in. by 

7. 6 oak beams each 3 by 4 -\ 

in. by 15 ft. 

8. 500 pieces of ^-in. square white pine cap strips each 

1 yd. long. 

9. A solid mahogany table top which is (j ft. in diameter 

and |- in. thick. 

10. The wood required for a floor 25 ft, by 15 ft. 6 in., if 

f-in. thick white pine 

is 

used. 

11. By means of a bar graph compare the weights of the 

metals in Table 5. 

12. Represent by a bar graph the weights 

given 

in Table 5. 

of the wood 

13. 50 round aluminum rods each 15 ft. long and f in. in 

diameter. 

14. Find the weight of the spruce I beam, shown in 

Fig. 92.

Board 


-U 

k- 

_ /2 ' 

-I 

LJ 

U'"4 

Fis. 92. I beam. 

15. Find the weight of 1,000 of each of the steel items 


Fig. 93. 

(b) 

Job 4: Board Feet 

Every mechanic sooner or later finds himself ready to 

purchase some lumber. In the lumberyard he must know 

I Booird foot 

I 

fool 

Fig. 94. 

the meaning of "board feet," because that is how most 

lumber is sold.


Definition: 

A board foot is a unit of measure used in lumber work. A 

board having a surface area of 1 sq. ft. and a thickness of 

1 in. or less is equal to 1 board foot (bd. ft.). 


Find the number of board feet in a piece of lumber 5 by 2 ft. by 

2 in. thick. 

Given : L 

Find: 

= 5 ft. 

W = 2ft. 

/ = 2 in. 

Number of board feet 

A =LX \v 

.1-5X2 

A = 10 sq. ft. 

Method: 

Board feet = .1 X t 

Board feet - 10 X 2 - 20 bd. ft. Ans. 

a. Find the surface area in square feet. 

/;. 

Multiply by the thickness in inches. 

Examples: 

1-3. Find the number of board feet in each of the pieces 

of rough stock shown in F'ig. 9.5. 

Example 2 Example 3 

Fig. 95.


4. Find the weight of each of the boards in Examples 1-3. 

5. Calculate the cost of 5 pieces of pine 8 ft. long by 

9 in. wide by 2 in. thick, at 11^ per board foot. 

6. Calculate the cost of this bill of materials: 

Job 5: 

Review Test 

1. Measure all dimensions on the airplane tail in Fig. 9(5, 

to the nearest 8 

Fig. 96. 

Horizontal stabilizers and elevators. 

2. Find the over-all length and height of the crankshaft 

in Fig. 97. 

Fig. 97. 

Crankshaft. 

3. Find the weight of the steel crankshaft in Fig. 97.

Volume and Weight 

79 

4. Find the area of the airplane wing in Fig. 98. 

49-3- 

^ 65-10"- -s 

Fi g 98. 

. 

5. Find the number of board feet and the weight of a 

spruce board 2 by 9 in. by 14 ft. long.

Chapter VI 

ANGLES AND CONSTRUCTION 

It has been shown that the length of lines can be measured 

by rulers, and that area and volume can be calculated with 

the help of definite formulas. Angles are measured with the 

Fig. 99. 

Protractor. 

help of an instrument called a protractor (Fig. 99). It will be 

necessary to have a protractor in order to be able to do any 

of the jobs in this chapter. 

(a) 

Fi 9 . 100. 

In Fig. 100(a), AB and BC are sides of the angle. B 

is called the vertex. The angle 

is known as LAEC or Z.CBA, 

since the vertex is 

always the middle letter. The symbol 

Z is mathematical shorthand for the word angle. Name

Angles and Construction 81 

the sides and vertex in Z.DEF', in Z.XOY. Although the 

sides of these three angles differ in length, yet 

Definition: 

An angle 

is 

the amount of rotation necessary to bring a 

line from an initial position to a final position. The length 

of the sides of the angle has nothing to do with the size 

of the angle. 

Job 1: How to Use the Protractor 


How many degrees does /.ABC contain? 

I A 

4ABC =70 

B 

Fi 3 . 101. 

Method: 

a. Place the protractor so that the straight edge coincides with 

the line BC (see Fig. 101). 

6. Place the center mark of the protractor on the vertex. 

c. Read the number of degrees at the point where line A B cuts 

across the protractor. 

d. Since /.ABC is less than a right angle, we must read the 

smaller number. The answer is 70.


Examples: 

1. Measure the angles in Fig. 102. 

-c s 

Fi 9 . 102. 

2. Measure the angles between the center lines of the 

parts of the truss member of the airplane 

rib shown in 

E 

Fig. 103. 

Fig. 1015. How many degrees are there in 

(a) 

(d) 

Z.AOB 

/.COA 

(I)} 

(e) 

LEOC 

LEOA 

(c) 

(/) 

LCOD 

Job 2: How to Draw an Angle 

The protractor can also be used to draw angles of a 

definite number of degrees, just as a ruler can be used to 

draw lines of a definite length.



Draw an angle of 30 

with A as vertex and with AB as one side. 

Method: 

a. Plaee the protraetor as if measuring an angle whose vertex 

is at A (see Fig. 104). 

4ABC = 30 

A ** 

Fig. 104. 

b. Mark a point such as (^ at the 80 graduation on the 

protractor. 

c. Aline from A to this point will make /.MAC = 30. 

Examples: 

Draw angles of 

1. 40 2. 60 3. 45 4. 37 6. 10 

6. 90 7. 110 8. 145 9. 135 10. 175 

11. With the help of a protractor bisect (cut in half) 

each of the angles in Examples, 

1 -5 above. 

12. Draw an angle of 0; of 180. 

13. Draw angles equal to each of the angles in Fig. 105. 

Y 

C 

Fig. 105. 

O


14. Draw angles equal to one-half of each of the angles in 

Fig. 105. 

Job 3: 

Units of Angle Measure 

So far only degrees have been mentioned in the measurement 

of angles. There are, however, smaller divisions than 

the degree, although only very 

skilled mechanics will have much 

occasion to work with such small 

units. Memorize the following 

table: 

TABLE 6. 

ANGLE MEASURE 

60 seconds (") 

= 1 minute (') 

GO minutes 1 degree () 

Questions: 

90 degrees 

Fig. 106. 18 degrees 

360 degrees 

1. How many right angles are there 

(a) in 1 straight angle? 

(ft) 

in a circle? 

2. How many minutes are there 

(a) in5? (6) in 45? (c) in 90? 

3. How many seconds are there 

(a) in 1 degree? 

(6) in 1 right angle? 

4. Figure 107 shows the position 

patch. Calculate the number of degrees in 

(a) /.DEC (I)} 

(c) ^FBC (d) 

(e) LAEF (f) 

Definition: 

An angle whose vertex is the 

center of a circle is 

called a central 

angle. For instance, Z.DBC in the 

= 1 right angle 

= 1 straight angle 

= 1 circle 

of rivets on a circular


circular patch in Fig. 107 is a central angle. Name any other 

central angles in the same diagram. 

Examples: 

1. In your notebook draw four triangles as shown in Fig. 

108. Measure as accurately as you can each of the angles in 

B A B B C 

each triangle. What conclusion do you draw as to the sum 

of the angles of any triangle? 

2. Measure each angle in the quadilaterals (4-sided 

plane figures) in Fig. 109, after drawing similar figures 

n 

Parallelogram 

Rectangle 

Square 

Trapezoid 

Fi g . 109. 

Irregular 

quadrilateral


in your own notebook. Find the sum of the angles of a 

quadilateral. 

Point 1 Point 2 

Fig. 110. 

3. Measure each angle in Fig. 110. Find the sum of the 

angles around each point. 

Memorize: 

1. The sum of the angles of a triangle is 180. 

2. The sum of the angles of a quadilateral is 300. 

3. The sum of the angles around a point is , CH)0. 

Job 4: Angles 

in Aviation 

This job will present just two of the many ways in which 

angles are used in aviation. 

A. Angle of Attack. The angle of attack is the angle 

between the wind stream and the chord line of the airfoil. 

In Fig. Ill, AOB is the angle of attack. 

is the angle ofattack 

Fig. 111. 

Wind 

Chord fine 

ofairfoil 

The lift of an airplane increases as the angle 

of attack is 

increased up to the stalling point, called the critical angle. 

Examples: 

1-4. Estimate the angle of attack of the airfoils in Fig. 

112. Consider the chord line to run from the leading edge


to the trailing edge. The direction of the wind is shown by 

W. 

5. What wind condition might cause a situation like the 

one shown in Example 4 Fig. 112? 

3. 

Fig. 112. 

B. Angle of Sweepback. Figure 113 shows clearly that 

the angle of sweepback is the angle between the leading 

edge and a line drawn perpendicular to the center line of 

the airplane. In the figure, Z.AOB is the angle of sweepback. 

The angle of 

sweepback 

is 4.AOB 

Fi 9 . 113. 

Most planes now being built have a certain amount of 

sweepback in order to help establish greater stability. 

Sweepback is even more important in giving the pilot an 

increased field of vision. 

Examples: 

Estimate the angle of sweepback of the airplanes in 

Figs. 114 and 115.

Mathematics for the Aviation Trades 

Fig. 114. 

Vultee Transport. (Courtesy of Aviation.) 

v^x- 

Fig. 1 1 5. Douglas DC-3. (Courtesy of Aviation.) 

Job 5: 

To Bisect an Angle 

This example has already been done with the help of a 

protractor. However, it is possible to bisect an angle with a 

ruler and a compass more accurately than with the protractor. 

Why? 

Perform the following construction in your notebook. 

ILLUSTRATIVE CONSTRUCTION 

Given : /.A BC 

Required To bisect : ,


Method: 

a. Place the point of the compass at B (see Fig. 116). 

b. Draw an arc intersecting BA at D, and BC at E. 

c. Now with D and E as centers, draw arcs intersecting at 0. 

Do not change the radius when moving the compass from D to E. 

d. Draw line BO. 

Check the construction by measuring /.ABO with the 

protractor. Is it equal to ^CBO? If it is, 

bisected. 

the angle has been 

Examples: 

In your notebook draw two angles as shown in Fig. 117. 

1. Bisect ÂOB and Z.CDE. Check the work with a 

protractor. 

2. Divide /.CDE in Fig. 117 into four equal parts. Check 

the results. 

C 

3. Is it possible to construct a right angle by bisecting a 

straight angle? Try it. 

Job 6: 

To Bisect a Line 

This example has already been done with the help of a 

rule. Accuracy, however, was limited by the limitations 

of the measuring instruments used. By means of the following 

method, any line can be bisected accurately without 

first measuring its length.

90 Mathematics for t/ie Aviation Tracks 

Method: 


Given: Line AB 

Required: To bisect AB 

a. 

Open a compass a distance which you estimate to be greater 

than one-half of AB (see Fig. 118). 

i 6. First with A as center then with B as 

| 

center draw arcs intersecting at C and D. 

h- Do not change the radius when moving the 

\ 

compass from A to K. 

^ 

c. Draw line CD cutting line AB at 0. 

ls ' 

' 

Check this construction by measuring 

AO with a steel rule. Is AO equal to OB ? If it is, line AB 

has been bisected. 

Definitions: 

Line CD is called the perpendicular bisector of the line AB. 

Measure Z.BOC with a protractor. Now measure Z.COA. 

Two lines are said to be perpendicular to each other when 

they meet at right angles. 

Examples: 

1. Bisect the lines in Fig. 119 after drawing them in your 

notebook. Check with a rule. 

(a)

Angles anc/ Construction 91 

4. Lay off a line 9 T-# in. long. Divide it into 8 equal parts. 

What is the length of each part 

? 

5. Holes are to be drilled on the fitting shown in Fig. 120 

so that all distances marked 

A are equal. Draw a line 

in. long, and locate the 

centers of the holes. Check 

the results with a rule. 

*"* 

-4 

A -)l(--A ~4*-A-->\*-A -* 

*"**' 

6. Draw the perpendicular 

bisectors of the sides of any triangle. Do they meet in one 

point ? 

l9 ' 

Job 7: 

To Construct a Perpendicular 

A. To Erect a Perpendicular at Any Point on a Line. 


Given: Line AB, and point P on line AB. 

Required: To construct a line perpendicular to AB at point P. 

Method: 

a. With P as center, using any convenient radius, draw an arc 

I \ c. Draw line OP. 

A C P D B 

Fig. 121. Check: 

cutting AB at C and D (see Fig. 121). 

b. First with C as center, then with D 

as center and with any convenient radius, 

draw arcs intersecting at 0. 

Measure LOPE with a protractor. Is it a right angle? If it 

then OP is 

perpendicular to AB. What other angle is 90? 

is, 

B. To Drop a Perpendicular to a Line from Any Point 

Not on the Line. 


Given: Line AB and point P not on line AB. 

Required: To construct a line perpendicular to AB and passing 

through P.


Method: 

With P as center, draw an arc intersecting line AB at C and D. 

Complete the construction with the help of Fig. 122. 

D 

Fig. 122. 

Examples: 

] 1. In your notebook draw any diagram similar to 

Fig. 123. Construct a perpendicular to line AB at point P. 

+C 

Al 

Fi 9 . 123. 

2. Drop a perpendicular from point C (Fig. 123) to line 

AB. 

Construct an angle of 

3. 90 4. 45 5. 2230 / 6. (>7i 

7. Draw line AB equal to 2 in. At A and B erect perpendiculars 

to AB. 

8. Construct a square whose side is 1^ in. 

9. Construct a right triangle in which the angles are 

90, 45, and 45. 

10-11. Make full-scale drawings of the layout of the 

airplane wing spars, in Figs. 

124 and 125. 

12. Find the over-all dimensions of each of the spars in 

Figs. 124 and 125.

Angles and Construction 

Job 8: 

To Draw an Angle Equal to a Given Angle 

This is an important job, and serves as a basis for many 

other constructions. Follow this construction in your notebook. 

Given: /.A. 


Required: To construct an angle equal to Z. ( with vertex at A'. 

Method: 

a. With A as center and with any convenient radius, draw arc 

KC (see Fig. 126a). 

A' 

(b) 

b. With the same radius, but with A f as the center, draw arc 

B'C' (see Fig. 1266).


c. With B as center, measure the distance EC. 

d. With B f as center, and with the radius obtained in (c), 

intersect arc B'C' at C'. 

e. Line A'C' will make /.C'A' B' equal to /.CAB. 

Check this construction by the use of the protractor. 

Examples: 

1. With the help of a protractor draw /.ABD and Z.EDB 

(Fig. 127) in your notebook, (a) Construct an angle equal to 

/.ABD. (V) Construct an angle equal to /.EDB. 

E 

Fig. 127. 

2. In your notebook draw any figures similar to Fig. 128. 

Construct triangle A'B'C'y each angle of which is 

equal 

to a corresponding angle of triangle ABC. 

3. Construct a quadrilateral A'B'C'D' equal angle for 

angle to quadrilateral ABCD. 

A C A D 

FiS. 128. 

Job 9: 

To Draw a Line Parallel to a Given Line 

Two lines are said to be parallel when they never meet, 

no matter how far they are extended. Three pairs of parallel 

lines are shown in Fig. 129. 

B ^ L N 

AC 

E 

G 

M P 

Fig. 129. AB is parallel to CD. EF is parallel to GH. LM is parallel to NP. 

H

Given: Line AH. 

Angles and Construction 


Required: To construct a line parallel to AB and passing 

through point P. 

Method: 

a. Draw any line PD through P cutting line AB at C (see Fig. 

130). 

_-jrvi 4' ^ 

Fig. 130. 

b. With P as vertex, construct an angle equal to /.DCB, as 

shown. 

c. PE is parallel to AB. 

Examples: 

1. lit your notebook draw any diagram similar to Fig. 131. 

Draw a line through C parallel to line A B. 

xC 

xD 

Fig. 131 

2. Draw lines through D and E, each parallel to line AB, 

in Fig. 131. 

3. Draw a perpendicular from E in Fig. 131 to line AB. 

4. Given /.ABC in Fig. 132. 

A 

Fig. 132.


a. Construct AD parallel to BC. 

b. Construct CD parallel to AB. 

What is the name of the resulting quadilateral? 

6. Make a full-scale drawing of this fitting shown in 

Fig. 133. 

|


3. Draw a line 7 in. long. Divide it into 6 equal parts. At 

each point of division erect a perpendicular. Are the 

perpendicular lines parallel to each other? 

(CL) 

(b) 

H 

ttapterVII 

GRAPHIC REPRESENTATION OF AIRPLANE 

DATA 

Graphic representation is constantly growing in importance 

not only in aviation but in business and government 

as well. As a mechanic and as a member of society, you 

ought to learn how to interpret ordinary graphs. 

There are many types of graphs: bar graphs, pictographs, 

broken-line graphs, straight-line graphs, and others. All of 

them have a common purpose: to show at a glance comparisons 

that would be more difficult to make from numerical 

data alone. In this case we might say that one picture 

is worth a thousand numbers. 

Origin* 

Horizon tot I ax is 

Fi 9 . 137. 

The graph is a picture set in a "picture frame/' This 

frame has two sides: the horizontal axis and the vertical 

axis, as shown in Fig. 137. These axes meet at a point called 

the origin. All distances along the axis are measured from 

the origin as a zero point. 

Job 1 : 

TTie Bar Graph 

The easiest way of learning how to make a graph is 

study the finished product carefully. 

98 

to

Graphic Representation of Airplane Data 99 

A COMPARISON OF THE LENGTH OF Two AIRPLANES 

DATA 

Scoile: space = 10 feet 

I 

Airplanes 

Fig. 1 38. (Photo of St. Louis Transport, courtesy of Curtiss Wright Corp.) 

The three steps in Fig. 

189 show how the graph in Fig. 

138 was obtained. Notice that the height of each bar may be 

approximated after the scale is established. 

Make a graph of the same data using a scale in which 1 

space equals 20 ft. Note how much easier it is to make a 

STEP 2 

Establish a convenient 

scale on each axis 

STEP 3 

Determine points on the 

scale from the data 

Airplanes 

I 2 

Airplanes 

Fig. 1 39. Steps in the construction of a bar graph. 

I 2 

Airplanes 

graph on "-graph paper" than on ordinary notebook paper. 

It would be very difficult to rule all the cross lines before 

beginning to draw up the graph.


Examples: 

1. Construct the graph shown in Fig. 140 in your own 

notebook and complete the table of data. 

A COMPARISON OF THE WEIGHTS OF FIVE MONOPLANES 

DATA 

Scoile I 

space * 1 000 Ib. 

Fig. 140. 

2. Construct a bar graph comparing the horsepower of 

the following aircraft engines: 

3. Construct a bar graph of the following data on the 

production of planes, engines, and spares in the United 

States : 

4. Construct a bar graph of the following data: 

Of the 31,264 pilots licensed on Jan. 1, 

were as follows: 

1940, the ratings


Job 2: Pictographs 

1,197 air line 

7,292 commercial 

988 limited commercial 

13,452 private 

8,335 solo 

Within the last few years, a new kind of bar graph called 

a piclograph has become popular. The pictograph does not 

need a scale since each picture represents a convenient 

unit, taking the place of the cross lines of a graph. 

Questions: 

1. How many airplanes does each figure in Fig. 141 

represent ? 

JanJ 

1935 

THE VOLUME OF CERTIFIED AIRCRAFT INCREASES STEADILY 

EACH FIGURE REPRESENTS 2,000 CERTIFIED AIRPLANES 

DATA 

1936 

1937 

1938 

1939 

1940 

Fig. 141. 

(Courtesy of Aviation.) 

2. How many airplanes would half a figure represent? 

3. How many airplanes would be represented by 3 

figures ? 

4. Complete the table of data.


5. Can such data ever be much more than approximate? 

Why? 

Examples: 

1. Draw up a table of approximate data from the 

pictograph, in Fig. 142. 

To OPERATE Quit CIVIL AIRPLANES WE II WE A (i ROWING FORCE OF PILOTS 

EACH FIGURE REPRESENTS 2,000 CERTIFIED PILOTS 

1937mum 

mommmmmf 

FiS. 142. (Courtesy of Aviation.) 

2. Do you think you could make a pictograph yourself? 

Try this one. Using a picture of a telegraph pole to represent 

each 2,000 miles of teletype, make a pictograph from 

the following data on the growth of teletype weather 

reporting in the United States: 

3. Draw up a table of data showing the number of 

employees in each type of work represented in Fig. 143.


EMPLOYMENT IN AIRCRAFT MANUFACTURING: 1938 

EACH FIGURE REPRESENTS 1,000 EMPLOYEES 

AIRPLANES 

DililljllllijyilllilllMjUiyililiJIIlli 

ENGINES 

INSTRUMENTS 

PROPELLERS 

PARTS & ACCES. 

III 

Fig. 143. 


4. Make a pictograph representing the following data 

on the average monthly pay in the air transport service: 

Job 3: 

The Broken-line Graph 

An examination of the broken-line graph in Fig. 144 will 

show that it differs in no essential way from the bar graph. 

If the top of each bar were joined by a line to the top of the 

next bar, a broken-line graph would result. 

a. Construct a table of data for the graph in Fig. 144. 

b. During November, 1939, 6.5 million dollars' worth of 

aeronautical products were exported. Find this point on 

the graph.

1 


c. What was the total value of aeronautical products 

exported for the first 10 months of 1939? 

EXPORT OF AMERICXX AEKON UTTICAL PRODUCTS: 1939 

DATA 

Jan. Feb. Mar. Apr May June July Auq.Sepi Och 

Fig. 144. 

Scale 

1 

space =$1,000,000 

Examples: 

1. Construct three tables of data from the graph in 

Fig. 145. This is really 3 graphs on one set of axes. Not 

varied from 

only does it show how the number of passengers 

PASSEXGEKS CARRIED BY DOMESTIC AIR LINES 

Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. 

Fi g . 145.

Graphic Representation of Airplane Data 1 05 

month to month, but it also shows how the years 1938, 1939, 

and 1940 compare in this respect. 

2. Make a line graph of the following data showing the 

miles flown by domestic airlines for the first 6 months of 

1940. The data are in millions of miles. 

3. Make a graph of the accompanying data on the number 

of pilots and copilots employed by domestic air carriers. 

Notice that there will have to be 2 graphs on 1 set of axes. 

Job 4: 

The Curved-line Graph 

The curved-line graph is generally used to show how two 

quantities vary with relation to each other. For example, 

the horsepower of an engine varies with r.p.m. The graph 

in Fig. 146 tells the story for one engine. 

The curved-line graph does not differ very much from the 

broken-line graph. Great care should be taken in the location 

of each point from the data. 

Answer these questions from the graph : 

a. What is the horsepower of the Kinner at 1,200 r.p.m. ? 

b. What is the horsepower at 1,900 r.p.m. ? 

c. At what r.p.m. would the Kinner develop 290 hp.?


d. What should the tachometer read when the Kinner 

develops 250 hp. ? 

CHANGE IN HORSEPOWER WITH R.P.M. 

KINNER RADIAL ENGINE 

DATA 

1000 1400 1800 

R. p.m. 

2200 

Fi g . 146. 

Vertical axis: 

f space = 25 hp. 

Horizontal axis' 

Ispace-ZOOr.p.m. 

e. 

Why isn't the zero point used as the origin for this 

particular set of data? If it were, how much space would be 

needed to make the graph ? 

Examples: 

1. Make a table of data from the graph in Fig. 147. 

280 

I 240 

o 

&200 

I 120 

< 

CD 

80 

CHANGE IN HORSEPOWER WITH R.P.M. 

RADIAL AIRPLANE ENGINE 

DATA 

^ 

1500 1600 1700 1800 1900 2000 2100 

R.p.m 

Fis. 147. 

R.p.m. 

1500 

1600 

1700 

1800 

1900 

2000 

2100 

B.hp. 

2. The lift of an airplane wing increases as the angle of 

attack is increased until the stalling angle is reached, 

Represent the data graphically.


Question: At what angle does the lift fall off? This is 

called the .stalling angle. 

3. The drag also increases as the angle of attack is 

increased. Here are the data for the wing used in Example 2. 

Represent this data graphically. 

Question: Could you have represented the data for 

Examples 2 and 3 on one graph? 

4. The lift of an airplane, as well as the drag, depends 

among other factors upon the area of the wing. The graph 

in Fig. 148 shows that the larger the area of the wing, 

the greater will be the lift and the greater the drag. 

a table of data 

Why are there two vertical axes? Draw up 

to show just how the lift and drag change with increased 

wing area.

108 Matfiemat/cs for the Aviation Trades 

LIFT AND DRAG VARY WITH WING AREA 

2 4 6 8 10 12 14 16 

Wing drea in square feet 

Fi 9 . 148. 


1. Make a bar graph representing the cost of creating a 

new type of aircraft (see Fig. 149). 

COST OF CREATING NEW OR SPECIAL TYPE AIRCRAFT 

BEECH 

AIRCRAFT CORP. 

ARMY TWIN 

Cost of First 

$180,000 

Ship 

Fig. 149. 

2. The air transport companies know that it takes a 

large number of people on the ground to keep their planes

Graphic Representation of Airplane Data 1 09 

in the air. Draw up a table of data showing how many 

employees of each type were working in 1938 (see Fig. 150). 

Am TRANSPORT'S ANNUAL EMPLOYMENT OF NONFLYING PERSONNEL: 1938 

EACH FIGURE REPRESENTS 100 EMPLOYEES 

U 

< 

v 

IAI 

OVERHAUL 

AND 

MAINTENANCE 

CREWS 

FIELD AND 

HANGAR CREWS 

DISPATCHERS 

turn 

STATION PERS. 

METEOROLOGISTS 

RADIO OPS. 

TRAFFIC PERS. 

OFFICE PERS. 

lift 

Fig. 1 50. (Courtesy of Aviation.) 

3. As the angle of attack of a wing is increased, both the 

and drag change as shown below in the accompanying 

table. Represent these data on one graph. 

4. The following graph (Fig. 151) was published by the 

Chance Vought Corporation in a commercial advertisement


to describe the properties of the Vought Corsair. Can you 

read it? Complete two tables of data: 

a. Time to altitude, in minutes: This table will show how 

many minutes the plane needs to climb to any altitude. 

PERFORMANCE OF THE VOUGHT-CORSAIR LANDPLANB 

Time to altitude, in minutes 

400 

4 

800 1200 1600 2000 2400 

Roite of climb at altitude, in feet per minute 

Fi 3 . 151. 

6. Rate of climb at altitude, in feet per minute: It is 

important 

to know just how fast a plane can climb at any altitude. 

Notice that at zero altitude, that is, at sea level, this 

plane can climb almost 1,600 ft. per min. How fast can it 

climb at 20,000ft.?

Part II 

THE AIRPLANE AND ITS 

WING 

Chapter VIII: The Wei 9 ht of the Airplane 

Job 1: Calculating \Ying Area 

Job 2: Mean Chord of a Tapered Wing 

Job 3: Aspect Ratio 

Job 4: The (iross Weight of an Airplane 

Job 5: Pay Load 

Job 6: Wing Loading 

Job 7: Power Loading 


Chapter IX: Airfoils and Wins Ribs 

Job 1: The Tipper ('amber 

Job 2: The Lower Camber 

Job 3: When the Data Are Given in Per Cent of Chord 

Job 4: The Nosepiece and Tail Seel ion 

Job 5: The Thickness of Airfoils 

Job 6: Airfoils with Negative Numbers 


111

CAapterVIII 

THE WEIGHT OF THE AIRPLANE 

Everyone has observed that a heavy transport plane has 

a much larger wing than a light plane. The reason is fairly 

simple. There is a direct relation between the area of the 

wing and the amount of weight the plane can lift. Here are 

some interesting figures: 

TABLE 7 

Draw a broken-line graph of this data, using the gross 

weight as a vertical axis and the wing area as a horizontal 

axis. What is the relation between gross weight and wing 

area ? 

Job 1: CalculatingWing Area 

The area of a wing is calculated from its plan form. Two 

typical wing-plan forms are shown in Figs. 152A and \5%B. 

The area of these or of any other airplane wing can be 

found by using the formulas for area that have already been 

learned. It is particularly easy to find the area of a rectangular 

wing, as in Fig. 153, if the following technical terms are 

remembered. 

113


Fig. 152A. Bellanca Senior Skyrocket with almost rectangular wins form. 


Fig. 152B. Douglas DC-2 with tapered wing form. (Courtesy of Aviation.) 

/- 

Trailing edge 

Wing 

A-Spcxn- 

Chord 

- 

f--- 

Wing 

- Leading edge 

Fig. 153. Rectangular wing.

The Weight of the Airplane 115 

Definitions: 

Span is the length of the wing from wing tip to wing tip. 

Chord is the width of the wing from leading edge to 

trailing edge. 

Formula: Area = span 

X chord 


Find the area of a rectangular wing whose span 

is 25.5 ft. and 

whose chord is 

4.5 ft. 

Given: Span = 25.5 ft. 

Chord = 4.5 ft. 

Find: 

Examples: 

Wing area 

Area = span X chord 

Area - 25.5 X 4.5 

Area = 114.75 sq. ft. 

Ans. 

1. Find the area of a rectangular wing whose span is 

20 ft. and whose chord is 4^ ft. 

2. A rectangular wing has a span of 36 in. and a chord of 

(> in. What is its area in square inches and in square feet? 

3. Find the area of (a) the rectangular wing in Fig. 154, 

(b) 

the rectangular wing with semicircular tips. 

* a 

I 

. 

35 > 6 

(a) 

(b) 

Fig. 154. 

7 

J 

4. Calculate the area of the wings in Fig. 155. 

Fig. 155.


6. Find the area of the tapered wing in Fig. 156. 

Fig. 156. 

V" J 

Job 2: Mean Cfcorc/ of a Tapered Wing 

From the viewpoint of construction, the rectangular wing 

form is 

probably the easiest to build. Why? It was found, 

however, that other types have better aerodynamical 

qualities. 

In a rectangular wing, the chord is the same at all points 

but in a tapered wing there is a different chord at each 

point (see Fig. 157). 

Wing span 

Definition: 

Fig. 1 57. A tapered wins h many chords. 

Mean chord is the average chord of a tapered wing. 

found by dividing the wing area by the span. 

It is 

area 

Formula: Mean chord 

span 


Find the mean chord of the Fairchild 45. 

Given: Area = 248 sq. ft. 

Span - 39.5 ft. 

Find: Mean chord 

area 

Chord = 

span 

248 

Chord = 

39.5 

Chord = 6.3 ft. Ans.

The Weight of the Airplane 

117 

Examples: 

1-3. Supply the missing data: 

Job 3: /Aspect Ratio 

Figures 158 and 159 show how a wing area of 360 sq. ft. 

might be arranged: 

Airplane 1: Span = 90 ft. 

Area = span X chord 

Chord = 4 ft. Area = 90 X 4 

Area = 360 sq. ft. 

Airplane 2: Span = 60 ft. 

Area = 60 X 6 

Chord - 6 ft, 

Area - 360 

Airplane 3: Span = sq. ft. 

30 ft. 

Area = 30 X 12 

Chord - 12 ft. Area = 360 sq. ft. 

Airplane 1: 

Fig. 159. 

It would be very difficult to build this wing 

strong enough to carry the normal weight of a plane. Why? 

However, it would have good lateral stability, which means 

it would not roll as shown in Fig. 1(>0. 

Airplane 2: These are the proportions of an average 

plane.

118 Mathematics tor the Aviation trades 

Fig. 160. An illustration of lateral roll. 

Airplane 3: This wing might have certain structural 

advantages but would lack lateral stability and good flying 

qualities. 

Aspect ratio is the relationship between the span and the 

chord. It has an important effect upon the flying characteristics 

of the airplane. 

Formula: Aspect r ratio -~- , , 

chord 

In a tapered wing, the mean chord can be used to find the 

aspect ratio. 

Examples: 


Find the aspect ratio of airplane 1 in Fig. 158. 

Given : 

Span = 90 ft. 

Chord - 4 ft. 

Find: 

Aspect ratio 

A ,. span 

Aspect ratio = 

, T 

chord 

Aspect ratio 

= ^f- 

Aspect ratio = 22.5 Ans. 

1. Complete the following table from the data supplied 

in Figs. 158 and 159.

TheWeight of the Airplane 

119 

2-5. Find the aspect ratio of these planes 

: 

6. Make a bar graph comparing the aspect ratios of the 

four airplanes in Examples 2-5. 

7. The NA-44 has a wing area of 255f sq. ft. and a span 

of 43 ft. (Fig. 161). Find the mean chord and the aspect 

ratio. 

Fig. 161. North American NA-44. (Courtesy of Aviation.) 

8. A Seversky has a span of 41 ft. Find its aspect ratio, if 

its wing area is 246.0 

sq. ft. 

Job 4: The GrossWeight of an Airplane 

The aviation mechanic should never forget that the 

airplane is a "heavier-than-air" machine. In fact, weight 

is such an important item that all specifications refer not 

only to the gross weight of the plane but to such terms as 

the empty weight, useful load, pay load, etc.

IZU 

Mathematics tor the Aviation trades 

Definition: 

Empty weight is the weight of the finished plane painted, 

polished, and upholstered, but without gas, oil, pilot, etc. 

Useful load is the weight of all the things 

that can be 

placed in the empty plane without preventing safe flight. 

This includes pilots, passengers, baggage, oil, gasoline, etc. 

Gross weight 

is the maximum weight that the plane can 

safely carry off the ground and in the air. 

Formula: Gross weight empty weight -f- useful load 

The figures for useful load and gross weight are determined 

by the manufacturer and U.S. Department of 

Fig. 162. The gross weight and center of gravity of an airplane can be found by 

this method. (Airplane Maintenance, by Younger, Bonnalie, and Ward.) 

Commerce inspectors. They should never be exceeded 

by the pilot or mechanic (see Fig. 162). 

Fig. 163. 

The Ryan SC, a low-wing monoplane. (Courtesy of Aviation.)


Examples: 


Find the gross weight of the Ryan S-C in Fig. 163. 

Given: Empty weight = 1,345 Ib. 

Find: 

Useful load = 805 Ib. 

Gross weight 

Gross weight = empty weight + useful load 

Gross weight = 1,345 Ib. + 805 Ib. 

Gross weight = 2,150 Ib. Ans. 

1-3. Calculate the gross weight of the planes in the 

following table: 

4-6. Complete the following table: 

7. Make a bar graph comparing the empty weights of 

the Beechcraft, Bennett, Cessna, and Grumman. 

Job 5: Pay Load 

Pay load is the weight of all the things that can be 

carried for pay, such as passengers, baggage, mail, and 

many other items (see Fig. 164). Manufacturers are always


Fig. 164. Pay load. United Airlines Mainliner being loaded before one of its 

nightly flights. (Courtesy of Aviation.) 

trying to increase the pay load as an inducement to buyers. 

A good method of comparing the pay loads of different 

planes is on the basis of the pay load as a per cent of the 

gross weight (see Fig. 164). 


The Aeronca model 50 two-place monoplane has a gross weight 

of 1,130 Ib. and a pay load of 210 Ib. What per cent of the gross 

weight is the pay load? 

Given: Pay load = 210 Ib. 

Gross weight = 1,130 Ib. 

Find: Per cent pay load 

Method: 

Per cent = 

Per cent 

pay load 

X 100 

gross weight 

100 

Per cent =* 18.5 Arts.


Examples: 

1. The monoplane in Fig. 165 is the two-place Akron 

Punk B, whose gross weight is 1,350 lb., and whose pay load 

Fig. 165. The Akron Funk B two-place monoplane. (Courtesy of Aviation.) 

is 210 lb. What per cent of the gross weight is the pay 

load? 

2-5. Find what per cent the pay load is of the gross 

weight in the following examples : 

6. Explain the diagram in Fig. 166. 

Fi 9 . 166.


Job 6: Wing Loading 

The gross weight of an airplane, sometimes tens of 

thousands of pounds, is carried on its wings (and auxiliary 

supporting surfaces) as surely as if they were columns of 

steel anchored into the ground. Just as it would be dangerous 

to overload a building 

till its columns bent, so it would 

be dangerous to overload a plane 

till 

safely hold it aloft. 

the wings could not 

Fig. 167. Airplane wings under static test. (Courtesy of Aviation.) 

Figure 167 shows a section of a wing under static test. 

Tests of this type show just how great a loading the structure 

can stand. 

Definition: 

Wing loading is 

the number of pounds of gross weight 

that each square foot of the wing must support in flight. 

Formula: Wing = ^ 

loading ; 

wing area 


A Stinson Reliant has a gross weight of 3,875 Ib. 

area of 258.5 sq. ft. Find the wing loading. 

Given: Gross weight 

= 3,875 Ib. 

Area = 258.5 sq. ft. 

and a wing


Find : 

Wing loadin g 

Examples: 

Wing loading = gross weight 

wing area 

Wing loading = 3,875 

258.5 

Wing loading = 14.9 Ib. per sq. ft. 

Ans. 

1. The Abrams Explorer has a gross weight of 3,400 Ib. 

and a wing area of 191 sq. ft. What is its wing loading? 

2-4. Calculate the wing loading 

following table: 

of the Grummans in the 

5. Represent by means of a bar graph the wing loadings 

and wing areas of the Grumman planes in the preceding 

table. One of these planes is shown in Fig. 168. 

Fig. 168. 

Grumman G-37 military biplane. (Courtesy of Aviation.)


6. The Pasped Skylark has a wing span 

of 35 ft. 10 in. 

and a mean chord of 5.2 ft. Find the wing loading 

gross weight is 

1,900 Ib. 

Job 7: Power Loading 

if the 

The gross weight of the plane must not only be held 

aloft by the lift of the wings but also be carried forward 

by the thrust of the propeller. A small engine would not 

provide enough horsepower for a very heavy plane; a large 

engine might "run away" with a small plane. The balance 

or ratio between weight and engine power 

is expressed by 

the power loading. 

Formula: Power loading 

= ?-- 

horsepower 


A Monocoupe 90A has a gross weight of 1,610 Ib. and is 

powered by a Lambert 90-hp. engine. What is the power loading? 

Given: Gross weight = 1,610 Ib. 

Find: 

Examples: 

Horsepower 90 

Power loading 

~ , ,. gross weight 

1 ower loading = ^-- 

horsepower 

T> i j- 

rower loading 

= 

90 

Power loading 

= 17.8 Ib. per hp. 


Arts.


Fig. 169. 

The Waco Model C. (Courtesy of Aviation.) 

Empty weisht = 2,328 Ib. Useful load = 1,472 Ib. Engine = 300 hp, 

170. The Jacobs L-6 7 cylinder radial engine used to power the Waco C 

(Courtesy of Aviation.)


4. Does the power loading increase with increased gross 

weight? Look at the specifications for light training planes 

and heavy transport planes. Which has the higher power 

loading? 

of a well- 

Note: The student may 

school or public library, or by obtaining a copy 

known trade magazine such as Aviation, Aero Digest, etc. 

find this information in his 

5. Find the gross weight and the power loading of the 

Waco model C, powered by a Jacobs L-6 7 cylinder radial 

engine (see Figs. 169 and 170). 


The following are the actual specifications of three 

different types of airplanes: 

1. Find the wing and power loading of the airplane in 

Fig. 171, which has the following specifications: 

Gross weight 

= 4,200 Ib. 

Wing area = 296.4 sq. ft. 

Engine 

= Whirlwind, 420 hp. 

Fig. 171. Beech Beechcraft D five-place biplane. (Courtesy of Aviation.) 

2. Find (a) the wing loading; (6) the power loading; (c) 

the aspect ratio; (d) the mean chord of the airplane in 

Fig. 172, which has the following specifications: 

Gross weight = 24,400 Ib. 

Wing area 

Engines 

Wing span 

987 sq. ft. 

= 2 Cyclones, 900 hp. each 

- 95 ft.


. 1 72. Douglas DC-3 24-placc monoplane. (Courtesy of Aviation.) 

3. Find (a) the gross weight; (6) the per cent pay load; 

(c) the mean chord; (d) the wing loading; (e) the power 

H 

28'S"' 

Fig. 173. Bellanca Senior Skyrocket, six-place monoplane. (Courtesy of Aviation.) 

loading; (/) the aspect ratio of the airplane in Fig. 173, 

which has the following specifications 

: 

Weight empty = 3,440 Ib. 

Useful load = 2,160 Ib. 

Pay load 

Wing area 

Engine 

Span 

986 Ib. 

= 359 sq. ft. 

= P. & W. Wasp, 550 hp. 

= 50 ft. 6 in. 

4. Does the wing loading increase with increased gross 

weight? Look up the specifications of six airplanes to prove 

your answer.

Chapter IX 

AIRFOILS AND WING RIBS 

The wind tunnel has shown how greatly the shape of the 

airfoil can affect the performance of the plane. The airfoil 

the manu- 

section is therefore very carefully selected by 

facturer before it is used in the construction of wing ribs. 

N.A.C.A.22 

N.A.C.A.OOI2 

Rib shape of 

Symmetrical 

Douglas DC3 

rib shape 

Fl g 174.- -Three types of . airfoil 

Clark Y 

Rib shape of 

Aeronca 

section. 

No mechanic should change this shape in any way. Figure 

174 shows three common airfoil sections. 

The process of drawing up the data supplied by the 

manufacturer or by the government to full rib size is important 

since any inaccuracy means a change in the plane's 

performance. The purpose of this chapter is to show how to 

draw a wing section to any size. 

Definitions: 

Datum line is the base line or horizontal axis (see Fig. 

175). 

Vertical^ 

axis 

Upper camber 

Trailing 

Leading 

edge 

edge '" 

Fig. 175. 

130 

> 

Lower camber 

Datum line

Airfoils and Wing Ribs 131 

Vertical axis is a line running through the leading edge 

of the airfoil section perpendicular to the datum line. 

Stations are points on the datum line from which measurements 

are taken up or down to the upper or lower camber. 

Upper camber is the curved line running from the leading 

edge to the trailing edge along the upper surface of the 

airfoil section. 

Lower camber is the line from leading edge to trailing edge 

along the lower surface of the airfoil section. 

The datum line (horizontal axis) and the vertical axis 

have already been defined in the chapter on graphic 

representation. As a matter of fact the layout of an airfoil 1 

is 

identical to the drawing of any curved-line graph from 

given data. The only point to be kept in mind is that there 

are really two curved-line graphs needed to complete the 

airfoil, the upper camber and the lower camber. These will 

now be considered in that order. 

Job 1 : 

The Upper Camber 

The U.S.A. 35B is a commonly used airfoil. The following 

data can be used to construct a 5-in. rib. Notice that the last 

station tells us how long the airfoil will be when finished. 

AIRFOIL SECTION: U.S.A. 35B 

Data in inches for upper camber only 

Airfoil 

section: U.S.A. 35 B 

I l'/ 

2 2 2'/ 2 3 3'/ 2 4 4'/ 2 

Fis. 176. 

1 The term "airfoil'* is often substituted for the more awkward phrase "airfoil 

section" in this chapter. Technically, however, airfoil refers to the shape of the 

wing as a whole, while airfoil section refers to the wing profile or rib outline.


Directions: 

Step 1. 

176). 

Draw the datum line and the vertical axis (see Fig. 

'/ 2 I l'/ 2 2 2'/2 3 3'/2 4 4'/2 5 

Fig. 177. 

Step 2. 

Step 3. 

Mark all stations as given in the data. 

At station 0, the data shows that the upper camber is 

fe in. above the datum line. Mark this point as shown in Fig. 177. 

Step 4. At station in., the upper camber is H in. above the 

datum line. Mark this point. 

Step 5. Mark all points in a similar manner on the upper 

camber. Connect them with a smooth line. The finished upper 

camber is shown in Fig. 178. 2'/2 3 3'/2 

Fig. 178. 

Job 2: 

The Lower Camber 

The data for the lower camber of an airfoil are always 

given together with the data for the upper camber, as shown 

AIRFOIL SECTION: U.S.A. 35B 

Fig. 179. 

in Fig. 179. In drawing the lower camber, the same diagram 

and stations are used as for the upper camber.


Directions: 

Step 1. At station 0, the lower camber is 7^ in. above the 

datum* line. Notice that this is the same point as that of the upper 

camber (see Fig. 180). 

SfepL 

Step 2'-* 

Fis. 180. 

Step 2. At station in., the lower camber is in. high, that is, 

flat on the datum line as shown in Fig. 180. 

Step 3. Mark all the other points on the lower camber and 

connect them with a smooth line. 

In Fig. 181 is 

one of the many planes using 

shown the finished wing rib, together with 

this airfoil. 

Notice that the 

airfoil 

Fig. 181. The Piper Cub Coupe uses airfoil section U.S.A. 35B. 

has more stations than you have used in your own 

work. These will be explained in the next few pages. 

Questions: 

1. Why does station have the same point on the upper 

and lower cambers?


2. What other station must have the upper and lower 

points close together? 

Examples: 

1-2. Draw the airfoils shown in Fig. 182 to the size 

indicated by 

the stations. 

Example 1. AIRFOIL SECTION: N-22 

All measurements are in inches. 

Example 2. 

AIRFOIL SECTION: N.A.C.A.-CYH 

N-22 NACA-CYH CLARK Y 

Fig. 182. 

The airfoil section N-22 is used for a wing rib on the 

Swallow; N.A.C.A.-CYH, which resembles the Clark Y 

very closely, 

is used on the Grumman G-37. 

3. Find the data for the section in Fig. 183 by measuring 

to the nearest 64th. 

Fls. 183.


4. Draw up the Clark Y airfoil section from the data in 

Fig. 184. 

AIRFOIL SECTION: 

CLARK Y 

The Clark V airfoil section is used in many planes, such 

as the Aeronca shown here. Note: All dimensions are 

in inches. 

Fig. 184. 

6. Make your own airfoil section, and find the data for it 

by measurement with the steel rule. 

Job 3: When the Data Are Given in Per Cent of Chord 

Here all data, including stations and upper and lower 

cambers, are given as percentages. This allows the mechanic 

to use the data for any rib size he wants; but he must first 

do some elementary arithmetic. 


A mechanic wants to build a Clark Y rib whose chord length 

is 30 in. Obtain the data for this size rib from the N.A.C.A. data 

given in Fig. 185. In order to keep the work as neat as possible 

and avoid any error, copy the arrangement shown in Fig. 185. 

It will be necessary to change every per cent in the N.A.C.A. 

data to inches. This should be done for all the stations and the 

upper camber and the lower camber.


AIRFOIL SECTION: CLARK Y, 80-iN. 

CHORD 

Fi g . 185. 

Stations: 

Arrange your work as follows, in order to obtain first the 

stations for the 30-in. rib. 

Rib Stations, Stations, 

Size, In. Per Cent In. 

30 X 0% = 30 X 0.00 = 

30 X 10% = 30 X 0.10 = 3.0 

30 X 20% = 30 X 0.20 = 6.0 

30 X 30% = 30 X 0.30 = 9.0 

30 X 40% = 30X0. 40= 12.0 

Calculate the rest of the stations yourself and fill in the proper 

column in Fig. 185.


Upper Camber: Arrange your work in a manner similar to the 

foregoing. 

Rib Upper Camber, Upper Camber, 

Size, In. Per Cent In. 

30 X 3.50% = 30 X .0350 = 1.050 

30 X 

. 

9.60% = 30 X .0960 = .880 

30 X 11.36% = 

Calculate the rest of 

the points on the upper camber. 

Insert these in the appropriate spaces in Fig. 185. Do the 

same for the lower camber. 

AIRFOIL SECTION: ("LARK Y, 30-iN 

CHORD 

Fig. 186. 

The data in decimals may be the final step before layout 

of the wing rib, providing a rule graduated in decimal parts 

of an inch is available.


If however, a rule graduated in ruler fractions is the only 

instrument available, it will be necessary to change the 

decimals to ruler fractions, generally speaking, accurate 

to the nearest 64th. It is suggested that the arrangement 

shown in Fig. 186 be used. Notice that the data in decimals 

are the answers obtained in Fig. 185. 

It is a good idea, at this time, to review the use of the* 

decimal equivalent chart, Fig. 64. 

Examples: 

1. Calculate the data for a 15-in. rib of airfoil N-22, and 

draw the airfoil section (see Fig. 187). 

N-22 SIKORSKY GS-M 

Fig. 187. 

2. Data in per cent of chord are given in Fig. 187 for 

airfoil Sikorsky GS-M. Convert these data to inches for a 

9-in. rib, and draw the airfoil section.


3. Draw a 12-in. diagram 

from the following data: 

of airfoil section Clark Y-18 

AIRFOIL SE( ITION: CLARK Y-18 

4. Make a 12-in. solid wood model rib of the Clark Y-18. 

Job 4: 

The Mosep/ece and Tail Section 

It has probably been observed that stations per cent 

and 10 per cent are not sufficient to give all the necessary 

S t a t i o n s 

01.252.5 5 7.5 10 20 

O 

Datum line 

Fig. 188. Stations between and 10 per cent of the chord. 

points for rounding out the nosepiece. As a result there are 

given, in addition to and 10 per cent, several more intermediate 

stations (see Fig. 188). 


Obtain the data in inches for a nosepiece of a Clark Y airfoil 

based on a 30-in. chord.


Here the intermediate stations are calculated exactly as before : 

Stations: 

Rib Size, Stations, Stations, 

In. Per Cent In. 

30 X 0% = 30 X =0 

30 X 1.25% = 30 X 0.0125 = 0.375 

30 X 2.5% = 30 X 0.025 = 0.750 

In a similar manner, calculate the remainder of the 

stations and the points on the upper and lower cambers. 

DATA 

I 

nches 

3 /4 l'/ a 2V 4 

Data based on 30 M chord 

Fig. 189. 

Noscpiec*: Clark Y airfoil section.


Figure 189 shows the nosepiece of the Clark Y airfoil 

section based upon a 30-in. chord. It is not necessary to lay 

out the entire chord length of 30 in. in order to draw up the 

nosepiece. Notice, that 

per 

1. The data and illustration are carried out only to 10 

cent of the chord or a distance of 3 in. 

2. The data are in decimals but the stations and points 

on the upper and lower cambers of the nosepiece were 

Note : All dimensions are in inches 

Fis. 190. Jig for buildins nosepiece of Clark V. 

located by using ruler fractions. Figure 190 is a blueprint 

used in the layout of a jig board for the construction of 

the nosepiece of a Clark Y rib. 

The tail section of a rib can also be drawn independently 

of the entire rib by using only part of the total airfoil data. 

Work out the examples without further instruction. 

Examples: 

All data are given in per cent of chord. 

1-2. Draw the nosepieces of the airfoils in the following 

tables for a 20-in. chord (see Figs. 191 and 192).


Fi g . 

20 40 60 80 

Per cent of chord 

191 .Section: N-60. 

AIRFOIL: N-60 

20 40 60 80 100 


Fig. 192. Section: U.S.A. 35A. 

AIRFOIL: U.S.A. 35A 

3-4. Draw the tail sections of the airfoils in the following 

tables for a 5-ft. chord. 

AIRFOIL: N-60 

AIRFOIL: U.S.A. 35A 

Job 5: 

The Thickness of Airfoils 

It has certainly been observed that there are wide 

variations in the thickness of the airfoils already drawn. 

The cantilever wing, which is braced internally, is more 

easily constructed if the thickness of the airfoil permits 

work to be done inside of it. A thick wing section also 

permits additional space for gas tanks, baggage, etc. 

On the other hand, a thin wing section has considerably 

less drag and is therefore used in light speedy planes.


to calculate the thickness of an airfoil 

It is 

very easy 

from either N.A.C.A. data in per cent of chord, or from the 

data in inches or feet. 

Since the wing rib is not a flat form, there is a different 

thickness at every station, and a maximum thickness at 

about one-third of the way back from the leading edge 

(see Fig. 193). 

Fig. 193. 


Find the thickness in inches of the airfoil I.S.A. 695 at all 

stations given in the data in Fig. 194. 

Method: 

To find the thickness of the airfoil at any station simply subtract 

the "lower" point from the " upper " point. Complete the 

table shown in Fig. 194 after copying it in your notebook. 

Examples: 

1. Find the thickness at all stations of the airfoil section 

in the following table, in fractions of an inch accurate to 

the nearest 64th. Data are given in inches for a 10-in. chord. 

AIRFOIL SECTION: 

U.S.A. 35B


2. Figure 193 shows an accurate drawing of an airfoil. 

Make a table of data accurate to the nearest 64th for this 

airfoil, so that it could be drawn from the data alone. 

3. Find the thickness of the airfoil in Fig. 193 at all 

stations, by 

actual measurement. Check the answers thus 

AIRFOIL SECTION: T.S.A. 695 

Fig. 194. 

obtained with the thickness at each station 

using the results of Example 2. 

obtained by 

Job 6: Airfoils with Negative Numbers 

It may have been noticed that thus far all the airfoils 

shown were entirely above the datum line. There are, 

however, many airfoils that have parts of their lower camber


below the datum line. This is indicated by the use of 

negative numbers. 

Definition: 

A negative number indicates a change of direction (see 

Fig. 195). 

+ 2 

-2 

Fig. 195. 

Examples: 

1. Complete the table in Fig. 

01 23456 

given in the graph. 

196 from the information 

Fi 9 . 196. 

2. Give the approximate positions of all points on both 

the upper and lower camber of the airfoil in Fig. 197. 

20 

-10 

10 20 30 40 50 60 70 80 90 100


Airfoil Section. N.A.C.A. 2212 is a good example of an 

airfoil whose lower camber falls below the datum line. Every 

point on the lower camber has a minus ( ) sign in front 

of it, except per cent which is neither positive nor negative, 

since it is right on the datum line (see Fig. 198). 

Notice that there was no sign in front of the positive 

numbers. A number is considered positive (+) unless a 

minus ( ) sign appears in front of it. 

In drawing up the airfoil, it has been stated that these 

per cents must be changed to decimals, depending upon the 

rib size wanted, and that sometimes it may be necessary to 

change the decimal fractions to ruler fractions. 

The methods outlined for doing this work when all 

numbers are positive (+), apply just as well when numbers 

are negative ( ). The following illustrative example will show 

how to locate the points on the lower camber only since all 

other points can be located as shown in previous jobs. 


Find the points on the lower camber for a 15-in. rib whose 

airfoil section is N.A.C.A. 2212. Data are given in Fig. 198. 

AIRFOIL SECTION: N.A.C.A. 2212 

Data in per cent of chord 

20 n 

20 40 60 80 


100 

Fis- 198. 

The Bell BG-1 uses this section. (Diagram of plane, courtesy of Aviation.)


Rib 

Camber, 

In. 

15 X 

Per Cent 

0% = 15 X 

15 

15 

15 

15 

X -1.46% = 15 X -0.0146 

X -1.96% = 15 X -0.0196 

X -2.55% = 15 X -0.0255 

X -2.89% = 

Size, 

Lower 

Lower 

Camber 

In. 

Lower 

Camber, 

Fractions 

= 

-0.2190 = -A 

-0.2940 = -if 

-0.3825 = -f 

The position of the points on the lower camber, as well 

as the complete airfoil, is shown in Fig. 198. 

Examples: 

1. Draw a 15-in. rib of the N.A.C.A. 2212 (Fig. 198). 

2. Draw the nosepiece (0-10 per cent) of the N.A.C.A. 

2212 for a 6-ft. rib. 

3. Find the data for a 20-in. rib of airfoil section N.A.C.A. 

4418 used in building the wing of the Gwinn Aircar (Fig. 

199). 



20 40 60 80 


100 

Fig. 199. 

The Gwinn Aircar uses this section.



1. Calculate the data necessary to lay out a 12-in. rib 

of the airfoil section shown in Fig. 200. All data are in per 

cent of chord. 

ZO 40 60 80 


Fig. 200. Airfoil section: Boeing 103. 

100 

AIRFOIL SECTION: BOEING 103 

2. Construct a table of data in inches for the nosepiece 

(0-15 per cent) of the airfoil shown in Fig. 201, based on a 

6-ft. chord. 

Fig. 201. 

20 40 60 80 


Airfoil section: Clark V-22. 

IOO


AIRFOIL SECTION: CLARK Y-22 

3. What is the thickness in inches at each station of a 

Clark Y-22 airfoil (Fig. 201) using a 10-ft. chord? 

4. Construct a 12-in. rib of the airfoil section N.A.C.A. 

2412, using thin sheet aluminum, or wood, as a material. 

This airfoil is used in constructing the Luscombe model 90 

(Fig. 202). 



20 40 60 80 


100 

Fig. 202. 

The Luscombe 90 uses this section. 

5. Construct a completely solid model airplane wing 

whose span is 15 in. and whose chord is 3 in., and use the 

airfoil section Boeing 103, data for which are given in 

Example 1.


Hint: Make a metal template of the wing 

as a guide (see Fig. 203). 

section to use 

Fis. 203. 

Wins-section template.

Part III 

MATHEMATICS OF MATERIALS 

Chapter X: Stren g th of Material 

Job 1 : Tension 

Job 2: Compression 

Job 3: Shear 

Job 4: Bearing 

Job 5: Required Cross-sectional Area 

Job (5: 

Review Test 

Chapter XI: Fittings, Tubing, and Rivets 

Job 1: Safe Working Strength 

Job 2: Aircraft Fillings 

Job 3: Aireraft Tubing 

Job 4: Aircraft Rivets 


Chapter XII: Bend Allowance 

Job 1: The Rend Allowance Formula 

Job 2: The Over-all Length of the Flat Pattern 

Job 3: When Inside Dimensions Are Given 

Job 4: When Outside Dimensions Are Given 


151

Chapter X 

STRENGTH OF MATERIALS 

"A study of handbooks of maintenance of all metal 

transport airplanes, which are compiled by the manufacturers 

for maintenance stations of the commercial airline 

operators, shows that large portions of the handbooks are 

devoted to detailed descriptions of the structures and to 

instructions for repair and upkeep of the structure. In 

handbooks for the larger airplanes, many pages of tables 

are included, setting forth the material t -^ 

of every structural part and the strenqth 

\-9 

7 c 

characteristic ol 

i i 

each item used. 

x*n 

What 

Bearing 

does this mean to the airplane mechanic ? 

Tension * Y////////////////////////A > 

Compression 

^ 

^j 

Bending 

I 

~ ' 

Shear 

*-r//^^


The purpose of this chapter is to explain the elementary, 

fundamental principles of the strength of materials. 

Job 1 : 

Tension 

A. Demonstration 1. Take three wires one aluminum, 

one copper, and one steel all ^V in. in diameter and 

suspend them as shown in Fig. 207. 

Fig. 206. 

Note that the aluminum wire will hold a certain amount 

of weight, let us say 

Fig. 207. 

2 lb., before it breaks; 

8/b. 

the copper will 

hold more than 4 lb.; and the 

steel wire will hold much more 

than either of the others. We can, 

therefore, say that the tensile 

strength of steel is greater than 

the tensile strength of either 

copper or aluminum. 

Definition: 

The tensile 

of an object 

is 

will support 

fails. 

strength 

the amount of weight it 

in tension before it 

The American Society for Testing Materials has used 

elaborate machinery to test most structural materials, and 

published their figures for everybody's benefit. These 

figures, which are based upon a cross-sectional area of 

1 sq. in., are called the ultimate tensile strengths (U.T.S.). 

Definition: 

Ultimate tensile strength is the amount of weight a bar I sq. 

in. in cross-sectional area will support in tension before it fails.

Strength of Materials 155 

TABLE 8. ULTIMATE TENSILE STRENGTHS 

(In Ib. per sq. in.) 

Aluminum 18,000 

Cast iron 20,000 

Copper 32,000 

Low-carbon steel 50,000 

Dural- tempered 55,000 

Brass 60,000 

Nickel steel 125,000 

High-carbon steel 175,000 

Examples: 

1. Name 6 stresses to which materials used in construction 

may be subjected. 

2. What is the meaning of * 

tensile strength? 

3. Define ultimate tensile 

strength. 

4. Draw a bar graph comparing 

the tensile strength of the 

materials in Table 8. 

B. Demonstration 2. Take 

? 

three wires, all aluminum, of 

aJ2/b. 

these diameters: ^2 in -> vfr in -> i 

and suspend them as shown 

in., 

in Fig. 208. 

Notice that the greater the 

diameter of the wire, the greater 

is 

128 Ib. 

Fi g 208. 

. 

its tensile strength. Using the data in Fig. 208, complete 

the following table in your own notebook.


Questions: 

1. How many times is the second wire greater than the 

strength? 

2. How many times is the third wire greater than the 

first in (a) cross-sectional area? (6) 

second in (a) area? (6) strength? 

3. What connection is there between the cross-sectional 

area and the tensile strength? 

4. Would you say that the strength of a material 

depended directly on its cross-sectional area ? Why 

? 

5. Upon what other factor does the tensile strength of a 

material depend? 

Many students think that- the length of a wire affects its 

tensile strength. Some think that the shape of the cross 

section is important in tension. Make up experiments to 

prove or disprove these statements. 

Name several parts of an airplane which are in tension. 

C. Formula for Tensile Strength. The tensile strength 

of a material depends on only (a) 

(b) 

ultimate tensile strength (U.T.S.). 

Formula: Tensile strength 

cross-sectional area (A)\ 

A X U.T.S. 

The ultimate tensile strengths 

substances can be found in Table 8, but the areas will in 

most cases require some calculation. 


for the more common 

Find the tensile strength of a f-in. aluminum wire. 

Given: Cross section: circle 

Diameter = | in. 

U.T.S. = 13,000 Ib. per sq. in. 

Find: 

a. Cross-sectional area 

b. Tensile strength 

a. Area = 0.7854 X D 2 

Area = 0.7854 X f X 1 

Area = 0.1104 sq. in. Ans.

Strength of Materials 1 57 

6. Tensile strength = A X U.T.S. 

Tensile strength = 0.1104 X 13,000 

Tensile strength 

- 1,435.2 Ib. Ans. 

Examples: 

1. Find the tensile strength of a ^j-in. dural wire. 

2. How strong 

is a |~ by 2^-in. cast-iron bar in tension? 

3. Find the strength of a i%-in. brass wire. 

4. What load will cause failure of a f-in. square dural 

rod in tension (sec Fig. 209) ? 

Fig. 209. Tie rod, square cross section. 

5. Find the strength in tension of a dural fitting at a 

point where its cross section is 

-^ by ^ in. 

6. Two copper wires are holding a sign. Find the greatest 

possible weight of the sign if the wires are each ^ in. 

in diameter. 

7. A load is being supported by four ^-in. nickel-steel 

the strength of this arrangement? 

8. A mechanic tried to use 6 aluminum ^2-in. rivets to 

bolts in tension. What is 

support a weight of 200 Ib. in tension. Will the rivets hold? 

9. Which has the greater tensile strength: (a) 5 H.C. 

steel ^r-in. wires, or (6) 26 dural wires each eV 

diameter? 

in. in 

10. What is the greatest weight that a dural strap 

H" by 3^-in. can support 

in tension? What would be the 

effect of drilling a ^-in, hole in the center of the strap? 

Job 2: 

Compression 

There are some ductile materials like lead, silver, copper, 

aluminum, steel, etc., which do not break, no matter how 

much pressure 

is 

put on them. If the compressive force is 

great enough, the material will become deformed (see 

Fig. 211).


On the other hand, if concrete or cast iron or woods of 

various kinds are put in compression, they will shatter 

C o repression 

Fi g . 210 

lead 

Due file 

material 

Fig. 211. 

into many pieces if too much load is applied. Think of what 

might happen to a stick of chalk in a case like that shown in 

Fig. 212. 10 Ib. 

iron 

Brittle 

material 

Fi 3 . 212. 

Definitions: 

Ultimate compressive strength (for brittle materials) is the 

number of pounds 1 sq. in. of the material will support in 

compression before it breaks. 

Ultimate compressive strength (for ductile materials) 

is the 

number of pounds 1 sq. in. of the material will support in 

compression before it becomes deformed. 

For ductile materials the compressive strength is 

to the tensile strength. 

equal


TABLE 9. 

ULTIMATE COMPRESSIVE STRENGTHS 


The formula for calculating compressive strength is 

very 

much like the formula for tensile strength : 

Formula: Compressive strength 

A X U.C.S. 

where A = cross section in compression. 

U.C.S. = ultimate compressive strength. 


Find the compressive strength of a \ by f in. bar of white pine, 

used parallel to the grain. 

Given: Cross section: rectangle 

L = | in., W = i in. 

U.C.S. = 5,000 Ib. per sq. in. 

Find: a. Cross-sectional area 

Examples: 

6. Compressive strength 

a. A = L X W 

A = f X t 

A = f sq. in. Arts. 

b. Compressive strength = A X U.C.S. 

Compressive strength = ^ 

X 5,000 

Compressive strength = 1,875 Ib. Ans. 

1. What is the strength in compression of a 3 by 5^-in. 

gray cast-iron bar? 

2. How much will a beam of white pine, 4 in. square, 

support if it is used (a) parallel to the grain? (b) across the 

grain ?


3. Four blocks of concrete, each 2 by 2 ft. in cross section, 

are used to hold up a structure. Under what load would 

they fail? 

4. What is the strength of a f-ft. round column of 

concrete ? 

(OL) 

m 

r" 

+--5 --H 

if*. 

(ct) 

Fig. 21 3. 

All blocks are of spruce. 

6. What is the strength of a 2-in. H.C. steel rod in 

compression? 

6. Find the strength in compression of each of the blocks 

of spruce shown in Fig. 213. 

Job 3: Shear 

Two plates, as shown in Fig. 215, have a tendency to 

cut a rivet just as a pair of scissors cuts a thread.


SHEAR 

Fig. 214. 

that is, 

The strength of a rivet, or any other material, in shear, 

its resistance to being cut, depends upon its crosssectional 

area and its ultimate shear strength. 

Formula: Shear strength 

= A X U.S.S. 

where A = cross-sectional area in shear. 

U.S.S. 

= ultimate shear strength. 

(M) 

Fi g . 215. 

TABLE 10. 

ULTIMATE SHEAR STRENGTHS 


The strength in shear of aluminum and aluminum alloy 

rivets is given in Chap. XI. 

Do these examples without any assistance from an 

illustrative example. 

Examples: 

1. Find the strength in shear of a nickel steel pin (S.A.E. 

2330) fk in. in diameter.


2. A chrome-molybdenum pin (S.A.E. X-4130) is -f in. in 

diameter. What is its strength in shear? 

3. What is the strength in shear of a -fV-in. brass rivet? 

4. A 2|- by 1^-in. spruce beam will withstand what 

maximum shearing load? 

5. What is the strength in shear of three ^-in. S.A.E. 

1015 rivets? 

Job 4: Bearing 

Bearing stress is a kind of compressing or crushing force 

which is met most commonly in riveted joints. It usually 

shows up by stretching the rivet hole and crushing the 

surrounding plate as shown in Fig. 216. 

o 

Original plate 

Fig. 216. 

Failure in bearing 

The bearing strength of a material depends upon 3 

factors: (a) the kind of material; (b) the bearing area; (c) 

the edge distance of the plate. 

The material itself, whether dural or steel or brass, will 

determine the ultimate bearing strength (U.B.S.), which is 

approximately equal to f times the ultimate tensile 

strength. 

TABLE 11. 

ULTIMATE BEARING STRENGTH 


Material 

U.B.S. 

Aluminum 18,000 

Dural-tempered 75,000 

Cast iron 100,000 

Low-carbon steel 75,000 

High-carbon steel 220,000 

Nickel steel 200,000


Bearing area is equal to the thickness of the plate multiplied 

by the hole diameter (see Fig. 217). 

Edge 

distance 

Diameter 

i Thickness 

where t 

Fig. 217. 

Formulas: Bearing area 

= thickness of plate. 

d = diameter of the hole. 

A = bearing area. 

t X d 

Bearing strength = A X U.B.S. 

U.B.S. = ultimate bearing strength. 

All the foregoing work is based on the assumption that 

the edge distance is at least twice the diameter of the hole, 

measured from the center of the hole to the edge of the 

plate. For a smaller distance the bearing strength falls off. 


Find the strength in bearing of a dural plate J 

yV-m- rivet hole. 

Given: t 

= | 

d = A 

Find: a. Bearing area 

6. Bearing strength 

a. Bearing area = t X d 

in. thick with a 

Examples: 

X T*V 

Bearing area = J 

Bearing area = xf^ sq. in. Ans. 

b. Bearing strength = A X U.B.S. 

Bearing strength = rls X 75,000 

Bearing strength = 1,758 Ib. Ans. 

1. Find the bearing strength of a |-in. cast-iron fitting 

with a |--in. rivet hole.


2. Find the bearing strength of a nickel-steel lug ^ in. 

thick which is drilled to carry a A-in. pin. 

3. What is the strength in bearing of a Tfr-in. dural plate 

with two A-in. rivet holes? Does the bearing strength 

depend upon the relative position 

of the rivet holes ? 

4. (a) What is the strength in bearing of the fitting in 

Fig. 218? 

Drilled^hole 

Fig. 218. 

(6) How many times is the edge distance greater than 

the diameter of the hole? Measure edge distance from the 

center of the hole. 

% Dural plate 

Hole drilled 

Wdiameter 

Fig. 219. 

5. Find the bearing strength from the dimensions given 

in Fig. 219. 

Job 5: Required Cross-sectional Area 

It has probably been noticed that the formulas for 

tension, compression, shear, and bearing are practically the 

same. 

Strength in tension = area X U.T.S. 

Strength in compression = area X U.C.S. 

Strength in shear = area X TJ.S.S. 

Strength in bearing 

= area X TJ.B.S. 

Consequently, instead of dealing with four different formulas, 

it is much simpler to remember the following: 

Formula: Strength 

= cross-sectional area X ultimate strength


In this general formula, it can be seen that the strength 

of a material, whether in tension, compression, shear, or 

bearing, depends upon the cross-sectional area opposing the 

stress and the ultimate strength of the kind of material. 

Heretofore the strength has been found when the 

dimensions of the material were given. For example, 

the tensile strength of a wire was found when its diameter 

was given. Suppose, however, that it is 

necessary to find the 

size (diameter) of a wire so that it be of a certain required 

strength, that is, able to hold a certain amount of weight. 

How was this formula obtained ? 

c ~ 

, ,. i strength required 

-1 

Formula: Cross-sectional area r-. ~, 

-rultimate 

strength 

It will be necessary to decide from reading the example 

what stress is 

being considered and to look up the right 

table before doing any work with the numbers involved. 


What cross-sectional area should a dural wire have in order to 

hold 800 Ib. in tension? 

Given: Required strength = 800 Ib. 

Find: 

Material = dural 

U.T.S. = 55,000 Ib. per sq. in. 

Cross-sectional area 

4 strength required 

Area = * -^ 

u 

. 

-r 

ultimate strength 

80 

5^000 

Area = 0.01454 sq. in. Arts. 

Check: t.s. = A X U.T.S. = 0.01454 X 55,000 = 799.70 Ib. 

Questions: 

1. Why doesn't the answer check exactly 

? 

2. How would you find the diameter of the dural wire? 

It is suggested that at this point the student review the 

method of finding the diameter of a circle whose area is 

given.


Examples: 

1. Find the cross-sectional area of a low-carbon steel 

wire whose required strength is 3,500 Ib. in tension. Check 

the answer. 

2. What is the cross-sectional area of a square oak beam 

Fig. 220. Tie rod, circular cross section. 

which must hold 12,650 Ib. 

grain ? 

in compression parallel 

to the 

3. What is the length of the side of the beam in Example 

2? Check the answer. 

4. A rectangular block of spruce used parallel to the 

grain must have a required strength in compression of 

38,235 Ib. If its width is 2 in., what is the cross-sectional 

length ? 

5. A copper rivet is required to hold 450 Ib. in shear. 

What is the diameter of the rivet? Check the answer. 

6. Four round high-carbon steel tie rods are required to 

hold a total weight of 25,000 Ib. What must be the diameter 

of each tie rod, if they are all alike? (See Fig. 220.) 


1. Find the tensile strength of a square high-carbon 

steel tie rod measuring -^ in. on a side. 

2 Holes 

drilled 

'/^radius 

Fig. 221.


2. Find the strength in compression of a 2 by 

1-in. oak 

beam if the load is applied parallel to the grain. 

3. What is the strength in shear of a steel rivet (S.A.E. 

1025) whose diameter is 3^ in.? 

4. What is the bearing strength of the fitting in Fig. 221 ? 

5. What should be the thickness of a f-in. dural strip in 

order to hold 5,000 Ib. in tension (see Fig. 222) 

? 

6. If the strip in Example 5 were 2 ft. 6 in. long, how 

much would it weigh?

Gupter XI 

FITTINGS, TUBING, AND RIVETS 

The purpose of this chapter is 

to apply 

the information 

learned about calculating the strength of materials to 

common aircraft parts such as fittings, tubing, and rivets. 

Job 1: SafeWorking Stress 

Is it considered safe to load a material until it is just 

about ready to break? For example, 

if a TS-ITL. low-carbon 

steel wire were used to hold up a load of 140 lb., would it be 

safe to stand beneath it as shown 

in Fig. 223? 

Fis, 223. Is this safe? 

Applying the formula tensile 

strength = A X TJ.T.S. will show 

that this wire will hold nearly 

145 lb. Yet it would not be safe 

to stand under it because 

1. This particular wire might 

not be so strong as it should be. 

2. The slightest movement of 

the weight or of the surrounding 

structure might break the wire. 

3. Our calculations might be 

wrong, in which case the weight 

might snap the wire at once. 

For the sake of safety, therefore, it would be wiser to use 

a table of safe working strengths instead of a table of 

ultimate strengths in calculating the load a structure can 

withstand. Using the table of ultimate strengths will 

168 

tell

Fittings, Tubing, and Rivets 169 

approximately how much loading a structure can stand 

before it breaks. Using Table 12 will tell the maximum loading 

that can be piled safely on a structure. 

TABLE 12. 

SAFE WORKING STRENGTHS 


Examples: 

1. What is the safe working strength of a i%-in. dural 

wire in tension ? 

2. What diameter H.C. steel wire can be safely used to 

hold a weight of 5,500 Ib.? 

3. A nickel-steel pin is required to withstand a shearing 

stress of 3,500 Ib. What diameter pin should be selected? 

4. What should be the thickness of a L.C. steel fitting 

necessary to withstand a bearing stress of 10,800 Ib., if the 

hole diameter is 0.125 in.? 

Job 2: Aircraft Fittings 

The failure of any fitting in a plane is very serious, and 

not an uncommon occurrence. This is sometimes due to a 

lack of understanding of the stresses in materials and how 

their strength is 

affected by drilling holes, bending operations, 

etc. 

Figure 224 shows an internal drag- wire fitting, just before 

the holes are drilled. 

Questions: 

1. What are the cross-sectional area and tensile strength 

at line BB 

1 

, using Table 12?


2. Two l-in- holes are drilled. What are now the cross- 

and area at line AA'? 

sectional shape 

3. What is the tensile strength at 

SAE I025 

L.C. STEEL 

D 

< 

Fis. 224. 

Examples: 

1. Using Table 12, find the tensile strength of the fitting 

in Fig. 225, (a) at section BB', (6) at section AA'. The 

' 

material is iV 

1 11 - low-carbon steel. 

2. Suppose that a ^-in. hole were drilled by a careless 

mechanic in the fitting in Fig. 225. What is the strength of 

this fitting in Fig. 226 now? 

Is this statement true: "The strength of a fitting is lowered 

by drilling holes in it"? 

Find the strength of the fittings in Fig. 227, using Table 12.

Fittings, Tubing/ and Rivets 171 

3 4iH.C. STEEL 

Example 3 

Example 4 

3 /64 

H 

S.A.E.I025 

Example 6 

3 /JS.A.E.I095 

Fig. 227 

Example 8 

Job 3: Aircraft Tubing 

The cross-sectional shapes of the 4 types of tubing most 

commonly used in aircraft work are shown in Fig. 228. 

Tubing is made either by the welding of flat stock or by 

cold-drawing. Dural, low-carbon steel, S.A.E. X-4130 or 

chrome-molybdenum, and stainless steel are among the 

Round Streamlined Square Rectangular 

Fig. 228. 

Four types of aircraft tubing. 

materials used. Almost any size, shape, 

or thickness can be 

purchased upon special order, but commercially the outside 

diameter varies from T\ to 3 in. and the wall thickness 

varies from 0.022 to 0.375 in.


A. Round Tubing. What is the connection between the 

outside diameter (D), the inside diameter (d} > 

and the wall 

thickness ()? 

Are the statements in Fig. 229 true? 

Fig. 229. 

Complete the following table : 

B. The Cross-sectional Area of Tubing. Figure 230 

shows that the cross-sectional area of any tube may be 

obtained by taking the area A of a solid bar and subtracting 

the area a of a removed center portion. 

D Minus 

[ 

a 

) 

d Equals 

S Minus a Equals 

Fig. 230.

Fittings/ Tubing, and Rivets 

Formula: Cross-sectional area = A 

For round tubes: A = 0.7854D 

2 

For square tubes: A = S 

2 

, 

a = 0.7854d 

2 

a 

a = s 

2 

. 

. 

173 

It will therefore be necessary to work out the areas of 

both A and a before the area of the cross section of a tube 

can be found. 


Find the cross-sectional area of a tube whose outside diameter 

is % in. and whose inside diameter is 

Given: D = 2 in. 

d = l 

Find: (1) A 

() a 

in. 

(3) Area of tube 

1^ in. 

(1) A = 0.7854 X D 2 

A = 0.7854 X 2 X 2 

(2) 

yl = 3.1416 sq. in. Ans. 

a is found in a similar manner 

a = 1.7667 sq. in. 

Ans. 

(3) Cross-sectional area == A a 

Cross-sectional area = 3.1416 - 1.7667 

Cross-sectional area = 1.3749 sq. in. 

Ans. 

Complete 

this table: 

The struts of a biplane are kept in compression, between 

the spars of the upper and lower wings, by means of the


tension in the bracing wires and tie rods. A few years ago 

almost all struts were of solid wooden form, but they are 

now being replaced by metal tubes. 

Answer the following questions because they will help to 

make clear the change from wood to metal parts in aircraft : 

1. What is the compressive strength of a round spruce 

strut whose diameter is 2^ in. ? 

2. What would be the strength of a dural strut of the 

same size and shape? 

3. Why are solid metal struts not used, since they are 

so much stronger than wooden ones ? 

a. If the spruce strut were 3 ft. long, what would it 

weigh ? 

b. What would the dural strut weigh ? 

4. Would a ^-in. H.C. steel round rod be as strong in 

compression as the spruce 

strut whose diameter is 2j 

in.? 

5. Why then are steel rods not used for struts ? 

Rods should never be used in compression because they 

will bend under a very small load. Tubing has great compressive 

strength compared to its weight. Its compressive 

strength can be calculated just like the strength of any 

other material. It should always be remembered, however, 

that tubing in compression will fail long before its full 

compressive strength is developed, because it will either 

bend or buckle. The length of a tube, compared to its 

diameter, is extremely important in determining the compressive 

load that the tube can withstand. The longer the 

tube the more easily it will fail. This fact should be 

kept in mind when doing the following examples. 

Examples: 

Use Table 12 in the calculations. 

1. Find the strength in compression of a S.A.E. 1015 

round tube, whose outside diameter is f in. and whose 

inside diameter is 0.622 in.

Fittings, Tubing, anc/ Rivets 1 75 

2. Find the strength of a square tube, S.A.E. 1025, 

whose outside measurement is 1^ 

thickness is O.OH3 in. 

in. and whose wall 

3. What is the strength in tension of a 16 gage round 

H.C. steel tube whose inside diameter is 

0.0930 in.? 

4. A nickel-steel tube, whose wall thickness is 0.028 in. 

and whose outside diameter is 1^ in., is placed in compression. 

What load could it carry before breaking if it did not 

bend or buckle? 

Job 4: Aircraft Rivets 

A. Types of Rivets. No study of aircraft materials would 

be complete without some attention to rivets and riveted 

joints. Since it is important that a mechanic be able to 

recognize each type of rivet, study Fig. 231 carefully, and 

notice that 

1. Most of the dimensions of a rivet, such as width of 

the head and the radius of the head, depend upon the 

diameter of the rivet, indicated by A in Fig. 231. 

2. The length of the rivet is measured from under the 

head (except in countersunk rivets) and is naturally 

independent of the diameter. 

Examples: 

1. Find all the dimensions for a button head aluminum 

rivet whose diameter is ^ in. (see Fig. 231). 

2. A countersunk head dural rivet has a diameter of 

Find the dimensions of the head. 

f in. 

3. Make a drawing, accurate to the nearest 32nd in., of a 

round head aluminum rivet whose diameter is f 

whose length 

is 2 in. 

in. and 

4. Make a drawing of a countersunk rivet whose diameter 

is -3- 

in. and whose length 

is 3 in. 

B. The Strength of Rivets in Shear. Many different 

kinds of aluminum alloys have been classified, and the


R- 

A 

5 

l**1. 

-j c 

^-76*^ 

* In sizes J in. and larger. f For sizes up to and including j^ in. diameter. 

Fig. 231. Common types of aluminum-alloy rivets. (From "The Riveting of 

Aluminum" by The Aluminum Co. of America.)

Fittings/ Tubing, and Rivets 177 

strength of each determined by 

direct test. The method of 

driving rivets also has an important effect upon strength 

as Table 13 shows. 

TABLE 13. 

STRESSES FOR DRIVEN KIVETS 

Examples: 

1. Find the strength in shear of a or-in. button head 

17S-T rivet, driven cold, immediately after quenching. 

2. What is the strength in shear of a ^-in. round head 

24S-T rivet driven cold immediately after quenching? 

3. Find the strength in shear of a f-in. flat head 2S rivet 

driven cold. 

4. Two 53S-W rivets are driven cold. What is their 

combined strength in shear, 

is -YQ in. ? 

if the diameter of each rivet 

5. Draw up a table of the shear strength of 2S rivets, 

driven cold, of these diameters: \ in., f- in., ^ in., f in., 

f in., n. 

C. Riveted Joints. There are two main classifications 

of riveted joints: lap and butt joints, as illustrated in 

Fig. 232. 

In a lap joint, the strength of the structure in shear is 

equal to the combined strength of all the rivets. In a butt 

joint, on the other hand, the shear strength of the structure 

is equal to the strength of the rivets on one side of the joint 

only. Why?


Y///////7// /3 K/ 

I 

I 

(a) 

(b) 

Fig. 232. Types of riveted joints: (a) lap joint/ (b) butt joint. 

Examples: 

1. Find the strength in shear of the lap joint in Fig. 233, 

using -- in. diameter 17S-T rivets driven hot. 

4> 23TE /////^W///A^ 

s. 233. 

Y///// 

Fis. 234. 

Lap joint, 5/64 in. 

diameter, 53S-T rivets, driven 

cold as received. 

Fig. 235. 

Double-plate butt joint, 

1 /64-in. steel rivets, driven hot. 

2. What is the strength of the butt joint in Fig. 233 if 

all rivets are ^4 -in. 24S-T driven cold immediately after 

quenching?

Fittings, Tubing, and Rivets 179 

3. What would be the strength in shear of a lap joint 

with one row of ten ITS -in. 2S rivets driven cold, as received ? 

4. Find the strength of the lap joint shown in Fig. 234. 

5. Find the strength of the butt joint shown in Fig. 235. 

Job 5: Review lest 

1. In a properly designed structure, no one item is disproportionately 

stronger or weaker than any other. Why? 

VM 

Fig. 236. Lap joint, dural plates, 3/16 in. diameter 17 S-T rivets, driven 

immediately after quenching. 

The riveted joint shown in Fig. 236 is 

in tension. Find 

(d) the ultimate strength in tension; (b) the strength of 

T 

L 

All maferia/s: High carbon $feel 

(a) (b) (c) 

Fig. 237. (a) Tie rod terminal; (b) clevis pin; (c) fitting. 

the rivets in shear; (c) the strength of the joint in bearing. 

If this joint were subjected to a breaking load, where would 

it break first? What changes might be suggested?


2. Examine the structure in Fig. 237 very carefully. 

Find the strength of (a) the tie rod terminal in tension; (6) 

the tie rod terminal in bearing; (c) the clevis pin in shear; 

(d) the fitting in tension; (e) the fitting in bearing. If the 

tie rod terminal were joined to the fitting by means of the 

clevis pin and subjected to a breaking load in tension, 

where would failure occur first ? What improvements might 

be suggested? 

NOTE: It will be necessary to find the ultimate strength 

in each of the parts of the above example.

V 

__ * ^- J 

I 

Chapter XII 

BEND ALLOWANCE 

A large number of aircraft factories are beginning to 

consider a knowledge of bend allowance as a prerequisite 

to the hiring of certain types of mechanics. Aircraft manufacturers 

in some cases have issued special instructions to 

their employees on this subject. 

\ Angle ofbend 

Many fittings require that 

Fig. 238. 

metal be bent from the flat 

piece, according to instructions given in blueprints or 

drawings. The amount of bending is measured in degrees 

and is called the angle of bend (see Fig. 238). 

R = Radius 

r*\ 

4J 

Good bend 

Fig. 239. 

Bad bend 

I 

When a piece of metal is bent, it is 

important 

to round 

he vertex of the angle of bend or the metal may break. A 

form is, 

therefore, used to assist the mechanic in making a 

good bend. The radius of this form as shown in Fig. 239 is 

called the radius of bend. 

181


A large radius of bend means a gradual curve; a very 

small radius means a sharp bend. Experience has shown 

that the radius of bend depends on the thickness of the 

metal. In the case of steel, for example, for cold bending, 

the radius of bend should not be smaller than the thickness 

of the metal. 

Job 1 : 

The Bend Allowance Formula 

This job 

is the basis of all the work in this chapter. Be 

sure it is understood before the next job is undertaken. 

Definition: 

Bend allowance (B.A.) is 

the length of the curved part of 

the fitting. It is practically equal to the length of an arc 

of a circle as shown in Fig. 40. 

Bend 

allowance 

Fis. 240. 

The amount of material needed for the bend depends 

upon the radius of bend (jR); the thickness of the metal 

(T}\ the angle of bend in degrees (N). 

Formula: B.A. - (0.0 

1 

743 X R + 0.0078 X T) X N 


Find the bend allowance for a f-in. steel fitting to be bent 90 

over a ^-in. radius, as shown in Fig. 241. 90* 

Fi S . 241.

fienc/ Allowance 1 8 3 

Given: R = \ in. 

T = | in. 

N = 90 

Find: B.A. 

Method: 

B.A. (0.01743 X fl + 0.0078 X T) X N 

B.A. = (0.01743 X i + 0.0078 X 1) X 90 

B.A. = (0.00872 + 0.00098) X 90 

B.A. = (0.00970) X 90 

B.A. = 0.8730 in. 

To the nearest 64th, in. Ans. 

a. Multiply, as indicated by the formula, within the 

parentheses. 

b. Add within the parentheses. 

c. 

Multiply the sum by the number outside the parentheses. 

Examples: 

1. Find the bend allowance for a Te-in. steel fitting to 

be bent 90 over a form whose radius is % in. 

2. What is the bend allowance needed for -J-in. dural 

fitting to be bent over a form whose radius is f in. to make 

a 45 angle of bend ? 

Fi 3 . 242. 

(b) 

3. A -g\-i n steel fitting is to be bent 00. The radius of 

bend is \ in. What is the bend allowance? 

4. Find the bend allowance for each of the fittings shown 

in Fig. 242. 

5. Complete the following table, keeping in mind that 

the thickness of the metal, R is the radius of bend, and 

T is 

that all dimensions are in inches.


BEND ALLOWANCE CHART 

(90 angle of bend) 

R 

t 

0.120 

005 

0.032 

Job 2: 

The Over-all Length of the Flat Pattern 

Before the fitting can be laid out on flat stock from a 

drawing or blueprint such as shown in Fig. 243 (a), it is 

N 

"BENT-UP"VIEW 

() 

Fis- 243. 

FLAT 

PATTERN 

(b) 

important to know the over-all or developed length of the 

flat pattern, which can be calculated from the bent-up 

drawing. If the straight portions of the fitting are called 

A and J5, the following formula can be used: 

Formula: Over-all length 

= A + B + B.A. 


Find the over-all length of the flat pattern in Fig. 244. Notice 

that the bend allowance has already been calculated. 

64 

Fi 9 . 244.

Given: 

Find: 

A ~ f in. 

B = 2^ in. 

B.A. = T& in. 

Over-all length 

Bend Allowance 185 

Examples: 

Over-all length = A + B + B.A. 

Over-all length = f + | + -& 

Over-all length = |f + 2^ + A 

Over-all length = 2fjr *n - 

Ans. 

1. Find the over-all length of a fitting where the straight 

parts are ^ in. and f in., if the bend allowance is -f$ in. 

2. Find the over-all length of the fittings shown in 

Fig. 245. 

Jr. 

k#--J 

(a) 

~T 

T 

IT 

s. 245. Fig. 246. 1 8-in. cold-rolled 

steel, 90 bend. 

3. Find the bend allowance and the over-all length of 

the fitting shown in Fig. 246. 

4. Draw the flat pattern for the fitting in Fig. 246, 

accurate to the nearest 64th of an inch. 

_ 

5. Draw the flat pattern for the fitting in Fig. 247, after 

finding the bend allowance and over-all length.


Job 3iWhen Inside Dimensions Are Given 

It is easy enough to find the over-all length when the 

exact length of the straight portions of the fitting are given. 

However, these straight portions must 

i 

^3 

Fig. 248. 

usually be found from other dimensions 

given in the drawing or blueprint. 

In this case, an examination of Fig. 248 

will show that the straight portion A is 

equal to the inner dimension d, minus the 

radius of bend jR. This can be put in terms 

of a formula, but it will be easier to solve 

each problem individually, than to apply a formula 

mechanically. 

Formula: A = d 

where A = length of one straight portion. 

d = inner dimension. 

R = radius of bend. 

B, which is the length of the other straight portion of the 

fitting, can be found in a similar manner. 

R 


Find the over-all length of the flat pattern for the fitting 


f 

Given: 

Find: 

Bend r 

90"J 

R = ? in. 

T - * in. 

N = 90 

B.A. = |i in. 

Over-all length 

A - 2* 

- | 

= 

B = If 

- i 

- 

Fig. 249. 

_i 

rfe,


Method: 

Over-all length = A + B + B.A. 

Over-all length = t + If + ft 

Over-all length = 4^ in. 

Ans. 

a. First calculate the length of the straight portions, A and B, 

from the drawing. 

b 

Then use the formula: over-all length = A + B + A.B. 

In the foregoing illustrative example, the bend allowance 

was given. Could it have been calculated, if it had not been 

given? How? 

Examples: 

1. Find the over-all length of the flat pattern 

of the fittings 

shown in Fig. 250. Notice that it will first be necessary 

to find the bend allowance. 

Bend 90 

A* 

J" 

(a) 

Fi 3 . 250. 

2. Find the bend allowance and over-all length of the 

flat patterns of the fittings in Fig. 251. Observe that in 

(a) one outside dimension is given. 

-is I*'*"" 

fa) 

251. 

(b) 

64 

3. Find the bend allowance and over-all length 

of the 

fitting shown in Fig. 252. Draw a full-scale flat pattern 

accurate to the nearest 64th.


Job 4: When Outside Dimensions Are Given 

In this case, not only the radius but also the thickness 

of the metal must be subtracted from the outside dimension 

in order to find the length of the straight portion. 

Formula: A = D R T 

where A = length of one straight portion. 

D = outer dimension. 

R = radius of bend. 

T = thickness of the metal. 

B can be found in a similar manner. 

Here again no formulas should be memorized. A careful 

analysis of Fig. 253 will show how the straight portion 

L -A 

jr> 

Fis. 253. 

of the fitting can be found from the dimensions given on 

the blueprint. 

Examples: 

1. Find the length of the straight parts of the fitting 


2. Find the bend allowance of the fitting in Fig. 254. 

3. Find the over-all length of the flat pattern 

fitting in Fig. 254. 

of the


0.049- 

Fig.254. 

3/64-in.L.C.rteel 

bent 90. 

Fig . 255. 

4. What is the over-all length of the flat pattern 

fitting shown in Fig. 255 ? The angle of bend is 90. 

of the 

r--/A-~, 

U-~// -> 

Fig. 256. 0.035 in. thickness, 2 bends of 90 each. 

6. Make a full-scale drawing of the flat pattern 

fitting shown in Fig. 256. 

of the 

Fig. 257. 1 /8-in. H.C. steel bent 90, 1 /4 in. radius of bend. 

6. What is thd over-all length of the fitting in Fig. 257? 


Figure 258 shows the diagram of a 0.125-in. low-carbon 

steel fitting. 

1. (a) Find the bend allowance for each of the three 

bends, if the radius of bend is ^ in. 

(6) Find the over-all dimensions of this fitting.


Bencl45 

..I 

2 i -r5 fle"QW 

AH dimensions 

are in inches 

Fig. 258. 

2. Find the tensile strength of the fitting in Fig. 258 if 

the diameter of all holes is \ in., (a) each end; (6) at the 

side having two holes. Use the table of safe working 

strengths page 169. 

3. Make a full-scale diagram of the fitting (Fig. 258), 

including the bend allowance. 

4. Calculate the total bend allowance and over-all length 

of the flat pattern for the fitting in Fig. 259. All bends are 

90

Part IV 

AIRCRAFT ENGINE MATHEMATICS 

Chapter XIII: Horsepower 

Job 1: Piston Area 

Job 2: Displacement of the Piston 

Job 3 : Number of Power Strokes 

Job 4: Types of Horsepower 

Job 5: Mean Effective Pressure 

Job 6: How to Calculate Brake Horsepower 

Job 7: The Prony Brake 


Chapter XIV: Fuel and Oil Consumption 

Job 1: Horsepower-hours 

Job 2: Specific Fuel Consumption 

Job 3: Gallons and Cost 

Job 4: Specific Oil Consumption 

Job 5: How Long Can an Airplane Stay Up? 


Chapter XV: Compression Ratio and Valve Timins 

Job 1 : 

Cylinder Volume 

Job 2: Compression Ratio 

Job 3: How to Find the Clearance Volume 

Job 4: Valve Timing Diagrams 

Job 5: How Long Does Each Valve Remain Open? 

Job 6: Valve Overlap 


191

Chapter XIII 

HORSEPOWER 

What is the main purpose of the aircraft engine? 

It provides the forward thrust to overcome the resistance 

of the airplane. 

What part of the engine provides the thrust? 

The rotation of the propeller provides the thrust (see 

Fig. 260). 

Fig. 260. 

But what makes the propeller rotate? 

The revolution of the crankshaft (Fig. 261) turns the 

propeller. 

What makes the shaft rotate? 

The force exerted by the connecting rod (Fig. 262) turns 

the crankshaft. 

What forces the rod to drive the heavy shaft around? 

The piston drives the rod. 

Trace the entire process from piston to propeller. 

It can easily be seen that a great deal of work is required 

to keep the propeller rotating. This energy comes from the 

burning of gasoline, or any other fuel, in the cylinder. 

193

1 94 Mat/iemat/cs for the Aviation Trades 

Fig. 261. Crankshaft of Wright Cyclone radial ermine. (Courtesy of Aviation.) 

Rg. 262. Connecting rods, Pratt and Whitney 

Aviation.) 

radial ensine. (Courtesy of

Horsepower 195 

In a very powerful engine, a great deal of fuel will be 

used and a large amount of work developed. We say such 

an engine develops a great deal of horsepower. 

In order to understand horsepower, we must first learn 

the important subtopics upon which this subject depends. 

Job 1 : 

Piston Area 

The greater the area of the piston, the more horsepower 

the engine will be able to deliver. It will be necessary to 

find the area of the piston before the horsepower 

engine can be calculated. 

of the 

Fig. 263. 

Piston. 

The top* of the piston, called the head, is 

known to be a 

circle (see Fig. 263). To find its area, the formula for the 

area of a circle is needed. 

Formula: A = 0.7854 X D 2 


Find the area of a piston whose diameter is 3 in. 

Given: Diameter = 3 in. 

Find : Area


A = 0.7854 X D 2 

A = 0.7854 X 3 X 3 

A = 7.0686 sq. in. Ans. 

Specifications of aircraft engines do not give the diameter 

of the piston, but do give the diameter of the cylinder, or 

the bore. 

Definition: 

Bore is equal to the diameter of the cylinder, but may 

correctly be considered the effective diameter of the piston. 

Examples: 

1. Find the area of the pistons whose diameters are 

6 in., 7 in., 3.5 in., 1.25 in. 

2-9. Complete the following: 

10. The Jacobs has 7 cylinders. What is its total piston 

area? 

11. The Kinner C-7 has 7 cylinders and a bore of 5f in. 

What is its total piston area?


12. A 6 cylinder Menasco engine has a bore of 4.75 in. 

Find the total piston area. 

The head of the piston may be flat, 

concave, or domed, 

as shown in Fig. 264, depending on how it was built by 

Effective 

piston 

Flat 

Concave 

S. 264. Three types of piston heads. 

Dome 

the designer and manufacturer. The effective piston area 

in all cases, however, can be found by the? method used in 

this job. 

Job 2: Displacement of the Piston 

When the engine is running, the piston moves up and 

down in its cylinder. It never touches the top of the cylinder 

Top center 

Bottom center 

Displacement 

Fig. 265. 

on the upstroke, and never comes too near the bottom 

of the cylinder on the downstroke (see Fig. 265).


Definitions: . 

' 

Top 

center is the 

highest point the piston 

reaches on its 

upstroke. 

Bottom center is the lowest point the head of the piston 

reaches on the downstroke. 

Stroke is the distance between top center and bottom 

measured in inches or in feet. 

center. It is 

Displacement is the volume swept through by the piston 

in moving from bottom center to top center. It is measured 

in cubic inches. It will depend upon the area of the moving 

piston and upon the distance it moves, that is, its stroke. 

Formula: Displacement area X stroke 


Find the displacement of a piston whose diameter is 6 in. and 

whose stroke is 5% in. Express the answer to the nearest tenth. 

Given: Diameter = 6 in. 

Find: 

Stroke 

= 5^ = 5.5 in. 

Displacement 

A = 0.7854 X Z> 2 

A = 0.7854 X 6 X 6 

A = 28.2744 sq. in. 

Disp. = A X S 

Disp. = 29.2744 X 5.5 

Disp. = 155.5 cu. in. 

Note that it is first necessary 

piston. 

Examples: 


Ans. 

to find the area of the

Horsepower 1 99 

4. The Aeronca E-113A has a bore of 4.25 in. and a 

stroke of 4 in. It has 2 cylinders. What is its total piston 

displacement? 

5. The Aeronca E-107, which has 2 cylinders, has a bore 

of 4-g- in. and a stroke of 4 in. What is its total cubic 

displacement? 

6. A 9 cylinder radial Wright Cyclone has a bore of 

6.12 in. and a stroke of 6.87 in. Find the total displacement. 

Job 3: 

Number of Power Strokes 

In the four-cycle engine the order of strokes is intake, 

compression, power, and exhaust. Each cylinder has one 

power stroke for two revolutions of the shaft. How many 

power strokes would there be in 4 revolutions? in 10 revolutions? 

in 2,000 r.p.m.? 

Every engine has an attachment on its crankshaft to 

which a tachometer, such as shown in Fig. 266, can be fas- 

Fig. 266. Tachometer. (Courtesy of Aviation.) 

tened. The tachometer has a dial that registers the number 

of revolutions the shaft is making in 1 minute. 

Formula: N 

- 

' 

X cylinders 

where N = number of power strokes per minute. 

R.p.m. 

* 

= revolutions per minute of the crankshaft.


A 5 cylinder engine is 


strokes does it make in 1 miri.? in 1 sec.? 

Given : 5 cylinders 

Find: 

1,800 r.p.m. 

N 

N = 

, . 

r r.p.m. , ,. , 

X cylinders 

making 1,800 r.p.m. How many power 

N = 4,500 power strokes per minute 

Ana. 

There are 60 sec. in 1 

min. 

Number of power strokes per second : 

N 4,500 

= 

. 

_ 

. 

= 75 

A 

power strokes per second Ans. 

Examples: 

1-7. Complete the following table in your own notebook: 

8. How many r.p.m. does a 5 cylinder engine make 

when it delivers 5,500 power strokes per minute?


9. A 9 cylinder Cyclone delivers 9,000 power strokes per 

minute. What is the tachometer reading? 

10. A 5 cylinder Lambert is tested at various r.p.m. as 

listed. 

Complete the following table and graph the results. 

Job 4: 

Types of Horsepower 

The fundamental purpose of the aircraft engine is 

to turn the propeller. This work done by the engine is 

expressed in terms of horsepower. 

Definition: 

One horsepower of work is equal to 33,000 Ib. being 

raised one foot in one minute. Can you explain Fig. 267? 

The horsepower necessary to turn the propeller is 

developed inside the cylinders by the heat of the combustion 

of the fuel. But not all of it ever reaches the propeller. 

Part of it is lost in overcoming the friction of the 

shaft that turns the propeller; part of it is used to operate 

the oil pumps, etc. 

There are three different types of horsepower (see Fig. 

268).


Definitions: 

Indicated horsepower (i.hp.) is the total horsepower 

developed in the cylinders. 

Friction horsepower (f.hp.) is that part 

horsepower that is used in overcoming 

of the indicated 

friction at the 

bearings, driving fuel pumps, operating instruments, etc. 

267. 1 hp. = 33,000 

ft.-lb. per min. 

268. Three types of horsepower. 

Brake horsepower (b.hp.) is the horsepower available to 

drive the propeller. 

Formulas: Indicated = horsepower brake horsepower -f- friction 

horsepower 

I.hp. 

= b.hp. -[- f.hp. 


Find the brake horsepower of an engine when the indicated 

horsepower is 45 and the friction horsepower 

Given: I.hp. 

= 45 

F.hp. = 3 

Find: B.hp. 

I.hp. = b.hp. + f.hp. 

is 3. 

45 = b.hp. + 3 

B.hp. = 42 Ans. 

Examples: 

1. The indicated horsepower of an engine is 750. If 

43 hp. is lost as friction horsepower, what is the brake 

horsepower ?

Horsepower 

203 


8. Figure out the percentage of the total horsepower 

that is used as brake horsepower in Example 7. 

This percentage 

is called the mechanical efficiency of the 

engine. 

9. What is the mechanical efficiency of an engine whose 

indicated horsepower is 

is 65? 

95.5 and whose brake horsepower 

10. An engine developes 155 b.hp. What is its mechanical 

efficiency if 25 hp. 

is lost in friction? 

Job 5: Mean Effective Pressure 

The air pressure all about us is approximately 15 Ib. per 

sq. in. This is also true for the inside of the cylinders before 

the engine is started; but once the shaft begins to turn, the 

pressure inside becomes altogether 

different. Read the following 

description of the 4 strokes of a 4-cycle engine very 

carefully and study Fig. 269. 

1. Intake: The piston, moving downward, acts like a 

pump and pulls the inflammable mixture from the carburetor, 

through the manifolds and open intake valve 

into the cylinder. When the cylinder 

is full, the intake valve 

closes.


During the intake stroke the piston moves down, making 

the pressure inside less than 15 Ib. per sq. in. This pressure 

is not constant at any time but rises as the mixture fills the 

chamber. 

2. Compression: With both valves closed and with a 

cylinder full of the mixture, the piston travels upward 

compressing the gas into the small clearance space above 

the piston. The pressure 

is raised by this squeezing of the 

mixture from about 15 Ib. per sq. 

per sq. in. 

in. to 100 or 125 Ib. 

(1) 

Intake 

stroke 

(2) 

Compression 

stroke 

(4) 

Exhaust 

stroke 

Fig. 269. 

3. Power: The spark plug supplies the light that starts 

the mixture burning. Between the compression and power 

strokes, when the mixture is 

compressed into the clearance 

space, ignition occurs. The pressure rises to 400 Ib. per 

sq. in. The hot gases, expanding against the walls of the 

enclosed chamber, push the only movable part, the piston, 

downward. This movement is transferred to the crankshaft 

by the connecting rod. 

4. Exhaust: The last stroke in the cycle 

is the exhaust 

stroke. The gases have now spent their energy in pushing 

the piston downward and it is necessary to clear the 

cylinder in order to make room for a new charge. The ex-


haust valve opens and the piston, moving upward, forces 

the burned gases out through the exhaust port and exhaust 

manifold. 

During 

the exhaust stroke the exhaust valve remains 

open. Since the pressure inside the cylinder is greater than 

atmospheric pressure, the mixture expands into the air. It 

Fig. 370. 

is further helped by the stroke of the piston. The pressure 

inside the cylinder naturally keeps falling off. 

The chart in Fig. 270 shows how the pressure changes 

through the intake, compression, power, and exhaust 

strokes. The horsepower of the engine depends upon the 

average of all these changing pressures. 

Definitions: 

Mean effective pressure is the average of the changing 

pressures for all 4 strokes. It will henceforth be abbreviated 

M.E.P. 

Indicated mean effective pressure 

is the actual average 

obtained by using an indicator card somewhat similar to 

the diagram. This is abbreviated I.M.E.P. 

Brake mean effective pressure 

is that percentage of the 

indicated mean effective pressure that is not lost in friction 

but goes toward useful work in turning the propeller. This 

is abbreviated B.M.E.P.


Job 6: How to Calculate Brake Horsepower 

We have already learned that the brake horsepower 

depends upon 4 factors: 

1. The B.M.E.P. 

2. The length of the stroke. 

3. The area of the piston. 

4. The number of power strokes per minute. 

Remember these abbreviations: 

Formula: B.hp. 

B.M.E.P. X L X A X N 

33,000 

iLLUSTRATIVE EXAMPLE 

Given : 

B.M.E.P. = 120 Ib. 

per sq. in. 

Stroke = 0.5 ft. 

Area = 50 sq. in. 

N = 3,600 per min. 

Find: Brake horsepower 

B.hp. = 

B.hp. = 

B.hp. = 327 


33,000 

120 X 0.5 X 50 X 3,600 

A ns. 

33,000 

It will be necessary to calculate the area of the piston 

and the number of power strokes per minute in most of the 

problems in brake horsepower. Remember that the stroke 

must be expressed in feet, before it is used in the formula.


Examples: 

1. Find the brake horsepower of an engine whose stroke 

is 3 ft. and whose piston area is 7 sq. in. The number of 

power strokes is 4,000 per min. and the B.M.E.P. is 

per sq. in. 

120 Ib. 

2. The area of a piston is 8 sq. in. and its stroke is 4 in. 

Find its brake horsepower 

if the B.M.E.P. is 100 Ib. per 

sq. in. This is a 3 cylinder engine going at 2,000 r.p.m. 

Hint: Do not forget to change the stroke from inches to 

feet. 

3. The diameter of a piston 

is 2 in., its stroke is 2 in., 

and it has 9 cylinders. When it is 

going at 1,800 r.p.m., the 

B.M.E.P. is 120 Ib. per sq. in. Find the brake horsepower. 

4-8. Calculate the brake horsepower of each of these 

engines: 

120 . 

1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 

R.p.m. 

Fig. 271. 

Graph of B.M.E.P. for Jacobs aircraft engine. 

9. The graph in Fig. 271 shows how the B.M.E.P. keeps 

changing with the r.p.m. Complete the table of data in your 

own notebook from the graph.


10. Find the brake horsepower of the Jacobs L-5 at each 

r.p.m. in the foregoing table, if the bore is 5.5 in. and the 

stroke is 5.5 in. This engine has 7 cylinders. 

Job 7: 

The Prony Brake 

In most aircraft engine factories, brake horsepower is 

calculated by means of the formula just studied. There are, 

in addition, other methods of obtaining it. 

Fig. 272. Prony brake. 

The Prony brake is 

built in many different ways. The 

flywheel of the engine whose power is to be determined is 

clamped by means of the adjustable screws between friction 

blocks. Since the flywheel tends to pull the brake in the 

same direction as it would normally move, it naturally 

pushes 

the arm downward. The force F with which it

pushes downward is 


=- 

measured on the scale in pounds. 

- 

, DL 2* X F X D X 

Formula: r.p.m. 

B.hp. 

F is the reading on the scale and is measured in pounds. 

This does not include the weight of the arm. 

D is the distance in feet from the center of the flywheel to 

the scale. 

TT may be used as 3.14. 


The scale of a brake dynamometer reads 25 Ib. when the shaft 

of an engine going 2,000 r.p.m. is 2 ft. from the scale. What is the 

brake horsepower? 

Given: F = 25 Ib. 

D = 2 ft. 

2,000 r.p.m. 

Find: B.hp. 

27r X F X D X r.p.m. 

B.hp. = 

B.hp. = 33,000 

2 X 3.14 X 25 X 2 X 2,000 

33,000 

Examples: 

B.hp. = 19.0 hp. 

Ans. 

1. The scale of a Prony brake 2 ft. from the shaft reads 

12 Ib. when the engine is going at 1,400 r.p.m. What is the 

brake horsepower of the engine? 

2. A Prony brake has its scale 3 ft. 6 in. from the shaft 

of an engine going at 700 r.p.m. What horsepower is 

developed when the scale reads 35 Ib. ? 

being 

3. The scale reads 58 Ib. when the shaft is 1 ft. 3 in. 

away. What is the brake horsepower when the tachometer 

reads 1,250 r.p.m.? 

4. It is important to test engines at various r.p.m. Find 

the brake horsepower of an engine at the following tachometer 

readings 

if the scale is 3 ft. from the shaft:


5. Graph the data in Example 4, using r.p.m. 

as the 

horizontal axis and brake horsepower as the vertical axis. 


1. Write the formulas for (a) piston area; (6) displacement; 

(c) number of power strokes; (d) brake horsepower. 

Fi3. 273. Szekeiy 3-cylinder air-cooled radial aircraft ensine. (Courtesy of 

Aviation.)


2. The following is part of the specifications on the Szekely 

aircraft engine (sec Fig. 273). Complete all missing data. 

Name of engine 

Szekely SR-3 model O 

Type 

3 cylinder, air-cooled, radial, overhead valve 

A.T.C No. 70 

B.M.E.P 

107 Ib. per sq. in. 

Bore 

Stroke 

Total piston area 

Total displacement 

Dept. of Commerce rating 

4J in. 

4J in. 

sq. in. 

cu. in. 

hp. at 1,750 r.p.m. 

3. Complete the missing data in the following specifications 

of the engine shown in Fig. 274 : 

Fig. 274. Menasco B-4 inverted in-line aircraft engine. (Courtesy of Aviation.) 

Name of engine Menasco B-4 

Type 4 cylinder in-line, inverted, air-cooled 

A.T.C No. 65 

Dept. of Commerce rating 

Manufacturer's rating 

B.M.E.P 

Total displacement 

Bore 

Stroke 

hp. at 2,000 r.p.m. 

hp. at 2,250 r.p.m. 


4^ in. 

51 in. 

cu. in.

Giapter XIV 

FUEL AND OIL CONSUMPTION 

Mechanics and pilots are extremely 

much gasoline and oil their engine will use, because aviation 

interested in how 

gasoline costs about 30ff a gallon, and an engine that wastes 

gasoline soon becomes too expensive to operate. That is 

why the manufacturers of aircraft list the fuel and oil 

consumption, in the specifications that accompany each 

engine. 

Fuel consumption is sometimes given in gallons per hour, 

or in miles per gallon as in an automobile. But both these 

methods are very inaccurate and seldom used for aircraft 

engines. The quantity of fuel and oil consumed keeps 

increasing as the throttle is opened and the horsepower 

increases. Also, the longer the engine 

is run, the more fuel 

and oil are used. 

We can, therefore, say that the fuel and oil consumption 

depends upon the horsepower of the engine and the hours of 

operation. 

Job 1 : 

Horsepower-hours 

Definition: 

The horsepower-hours show both the horsepower and 

running time of the engine in one number. 

Formula: Horsepower-hours 

= horsepower X hours 

where horsepower = horsepower of the engine. 

hour = length of time of operation 

212 

in hours.

Fuel and Oil Consumption 213 


A 65-hp. engine runs for hr. What is the number of horse- 

Given: 65 hp. 

power-hours ? 

Find: Hp.-hr. 

Hp.-hr. = hp. X hr. 

Examples: 

Hp.-hr. = 65 X 2 

Hp.-hr. = 130 Am. 

1. A 130-hp. engine 

is run for 1 hr. What is the number 

of horsepower-hours ? Compare your answer with the answer 

to the illustrative example above. 

2. A 90-hp. Lambert is run for 3 hr. 30 min. What is the 

number of horsepower-hours? 

3-9. Find the horsepower-hours for the following engines : 

Job 2: Specific Fuel Consumption 

A typical method of listing fuel consumption is in the 

number of pounds of fuel consumed per horsepower-hour, 

that is, the amount consumed by each horsepower for 1 hr. 

For instance a LeBlond engine uses about ^ Ib. of gasoline 

to produce 1 hp. for 1 

hr. We therefore say that the specific 

fuel consumption of the LeBlond is ^ Ib. per hp.-hr. This can 

be abbreviated in many ways. A few of the different forms 

used by various manufacturers follow:

214 Mat/iemat/cs for the Aviation Trac/es 

lb. per BHP per hour .50 lb. per HP. hour 

lb. /BHP /hour 

lb. /BHP-hour 

.50 lb. per HP.-hr. 

0.50 lb. /hp. hr. 

For the sake of simplicity the form lb. per hp.-hr. 

o.70i i i i i \ i i i 

used 

jo.65 

1300 1500 1700 1900 2100 

Revolutions per minute 

Fig. 275. Specific fuel consumption of 

will be 

for the work in this 

chapter. 

The specific fuel consumption 

changes 

with the 

r.p.m. The graph in Fig. 

275 shows that there is a 

different specific fuel con- 

the Menasco B-4. 

sumption at each throttle 

setting. As the number of revolutions per minute of the 

crankshaft increases, the specific fuel consumption changes. 

Complete the following table of data from the graph : 

Questions: 

1. At what r.p.m. 

is it most economical to operate the 

engine shown in Fig. 276, as far as gasoline consumption is 

concerned ?

Fuel and Oil Consumption 21 5 

2. The engine 

is rated 95 hp. at 2,000 r.p.m. What is the 

specific fuel consumption given in specifications for this 

horsepower? 

40 eo '; 

Fig. 276. Fuel sage. (Courtesy of Pioneer Instrument Division of Bendix 

' 

Aviation Corp.) 

Job 3: 

Gallons and Cost 

The number of pounds of gasoline an engine will consume 

can be easily calculated, if we know (a) the specific consumption; 

(6) the horsepower of the engine; (c) the running 

time. 

Formula: Total consumption specific consumption X horsepowerhours 


A Lycoming 240-hp. engine runs for 3 hr. Its specific fuel consumption 

is 0.55 Ib. per hp.-hr. How many pounds of gasoline will 

it consume? 

Given: 240 hp. for 3 hr. at 0.55 Ib. per hp.-hr. 

Find: 

Total consumption 

Total = specific consumption X hp.-hr. 

Total = 0.55 X 7-20 

Total = 396 Ib. 

Ans. 

Hint: First find the horsepower-hours. 

Examples: 

1-9. Find the total fuel consumption in pounds 

of the following engines: 

of each


10. Find the total consumption in pounds for the Whirlwind 

in Examples 8 and 9. 

If the weights that airplanes must carry in fuel only seem 

amazing, consider the following: 

The Bellanca Transport carries 1,800 Ib. of fuel 

The Bellanca Monoplane carries 3,600 Ib. of fuel 

The Douglas DC-2 carries 3,060 Ib. of fuel 

Look up the fuel capacity of 5 other airplanes and compare 

the weight of fuel to the total weight of the airplane. 

You now know how to find the number of pounds of 

gasoline the engine will need to operate 

for a certain 

number of hours. But gasoline 

is 

bought by the gallon. How 

many gallons 

will be needed? How much will it cost? 

One gallon of aviation gasoline weighs 6 Ib. and costs 

about 30ff. 


A mechanic needs 464 Ib. of gasoline. How many gallons should 

he buy at 30^f a gallon? How much would this cost? 

Given: 464 Ib. 

Find: (a) Gallons 

(6) Cost 

(a) 

A|A = 77.3 gal. 

(b) 77.3 X .30 = $3.19 Ana.

Fue/ and Oil Consumption 21 7 

Method: 

6. 

To get the number of gallons, divide the number of pounds by 

Examples: 

1. A mechanic needs 350 Ib. of gasoline. How many 

gallons does he need? If the price is 28^ per gallon, what 

is the cost? 

2. The price of gasoline is 20jS per gallon, and a mechanic 

needs 42 Ib. How much should he pay? 

3. A pilot stops at three airports and buys gasoline three 

different times : 

La Guardia Airport, 40 Ib. at 30^ per gallon. 

Newark Airport, 50 Ib. at 28ji per gallon. 

Floyd Bennett Field, 48 Ib. at 29^ per gallon. 

Find the total cost for gasoline on this trip. 

4. Do this problem without further explanation: 

A Bellanca Transport has a Cyclone 650-hp. engine 

whose specific fuel consumption is 0.55 Ib. per hp.-hr. On a 

runs for 7 hr. Find the number 

trip to Chicago, the engine of gallons of gasoline needed and the cost of this gasoline at 

25 per gallon. 

Note: The assumption here is that the engine operates at 

a constant fuel consumption for the entire trip. 

entirely true? 

Is this 

Shop Problem: 

What is meant by octane rating? Can all engines operate 

efficiently using fuel of the same octane rating? 

Job 4: Specific Oil Consumption 

The work in specific oil consumption is very much like the 

work in fuel consumption, the only point of difference being 

in the fact that much smaller quantities of oil are used.


The average specific fuel consumption is 0.49 Ib. per 

hp.-hr. 

One gallon of gasoline weighs 6 Ib. 

The average specific oil consumption is 0.035 Ib. per 

hp.-hr. 

One gallon of oil weighs 

7.5 Ib. 

Examples: 

Do the following examples by yourself: 

1. The Hornet 575-hp. engine has a specific oil consumption 

of 0.035 Ib. per hp.-hr. and runs for 3 hr. How many 

pounds of oil does it consume? 

2. A 425-hp. Wasp runs for 2-g- hr. If its specific oil consumption 

is 0.035 Ib. per hp.-hr., find the number of 

pounds of oil it uses. 

3-7. Find the weight and the number of gallons of oil 

used by each of the following engines 

: 

8. ALeBlond 70-hp. engine has a specific oil consumption 

of 0.015 Ib. per hp.-hr. How many quarts of oil would be 

used in 2 hr. 30 min. ? 

Job 5: How Long Can an Airplane Stay Up? 

The calculation of the exact time that an airplane can 

fly nonstop is not a simple matter. It involves consideration 

of the decreasing gross weight of the airplane due to the 

consumption of gas6line during flight, changes in horsepower 

at various times, and many other factors. However,


the method shown here will give a fair approximation 

of the answer. 

If nothing goes wrong with the engine, the airplane will 

stay aloft as long as there is gasoline left to operate the 

engine. That depends upon (a) the number of gallons of 

gasoline in the fuel tanks, and (6) the amount of gasoline 

used per hour. 

The capacity of the fuel tanks in gallons is always given 

in aircraft specifications. An instrument such as that appearing 

in Fig. 276 shows the number of gallons 

the tanks at all times. 

- is*.. ,. gallons in fuel tanks 

Formula: Cruising time = 

. 

jp- j 

gallons per hour consumed 

An airplane is 


of fuel in 

powered with a Kinner K5 which uses 8 gal. 

per hr. How long can it stay up, if there are 50 gal. 

tanks ? 

Examples: 

~ . . . 

gallons in fuel tanks 

Cruising time = -- 

n ? ; 

gallons per hour consumed 

Cruising time = %- = 6^ hr. 

^4rw. 

of fuel in the 

1. An Aeronca has an engine which consumes gasoline 

at the rate of 3 gal. per hr. How long can the Aeronca stay 

up, if it started with 8 gal. 

of fuel? 

2. At cruising speed an airplane using a LeBlond engine 

consumes 4f gal. per hr. How long can this airplane fly at 

this speed, if it has 12^ gal. 

of fuel in its tanks? 

3. The Bellanca Airbus uses a 575-hp. engine whose fuel 

consumption is 0.48 Ib. per hp.-hr. How long can this 

airplane stay up if its fuel tanks hold 200 gal.? 

4. The Cargo Aircruiser uses a 650-hp. engine whose 

consumption is 0.50 Ib. per hp.-hr. The capacity of the tank 

is 

150 gal. How long could it stay up? 

5. A Kinner airplane powered with a Kinner engine has 

50 gal. of fuel. When the engine operates at 75 hp., the


specific consumption is 0.42 Ib. per hp.-hr. How long could 

it fly? 

6. An airplane has a LeBlond 110-hp. engine whose 

specific consumption is 0.48 Ib. per hp.-hr. If only 10 gal. 

of gas are left, how long can it run ? 

7. A large transport airplane is lost. It has 2 engines of 

715 hp. each, and the fuel tanks have only 5 gal. altogether. 

If the lowest possible specific fuel consumption is 

0.48 Ib. per hp.-hr. for each engine, how long can the 

airplane stay aloft? 


1. A Vultee is 

powered by an engine whose specific fuel 

consumption is 0.60 Ib. per hp.-hr. and whose specific oil 

Fig. 277. 

Vultee military monoplane. (Courtesy of Aviation.) 

consumption is 0.025 Ib. per hp.-hr. when operating at 

735 hp. (see Fig. 277). 

a. How many gallons of gasoline would be used in 2 hr. 

15 min.? 

b. How many quarts of oil would be consumed in 1 hr. 

20 min. ? 

c. The fuel tanks of the Vultee hold 206 gal. How long can 

the airplane stay up before all the tanks are empty, 

operates continuously at 735 hp. ? 

if it 

d. The oil tanks of the Vultee have a capacity of 15 gal. 

How long would th engine operate before the oil tanks 

were empty?

fuel and Oil Consumption 221 

2. The Wright GR-2600-A5A is a 14 cylinder staggered 

radial engine whose bore is 6^ 

6fV in. When operating at 2,300 r.p.m. 

in. and whose stroke is 

its B.M.E.P. 

is 168 Ib. per sq. in. At rated horsepower, this engine's 

Fig. 278. 

Performance curves: Ranger 6 aircraft engine. 

specific fuel consumption is 0.80 Ib. per hp.-hr. Find how 

many gallons of gasoline will be consumed in 4 hr. 

Hint: First find the horsepower of the engine. 

3. The performance curve for the Ranger 6 cylinder, 

in-line engine, shown in Fig. 278, was taken from company 

Complete the following table of data from 

specifications. 

this graph:


A photograph of a Ranger 6 cylinder 

engine is shown in Fig. 279. 

inverted in-line 

Fig. 279. 

Ranger 6 cylinder in-line, inverted, air-cooled, aircraft engine. (Courtesy 

of Aviation.)

222 Mathematics for t/ie Aviation Trades 

4. Complete the following tables and represent 

by a line graph for each set of data. 

FUEL CONSUMPTION OF THE RANGER 6 

the results 

Aircraft engine performance curves generally show two 

types of horsepower: 

1. Full throttle horsepower. This is the power that the 

engine can develop at any r.p.m. Using the formula for b.hp. 

(Chap. XIII) will generally give this curve. 

2. Propeller load horsepower. This will show the horsepower 

required to turn the propeller at any speed.


A photograph of a Ranger 6 cylinder 

engine is shown in Fig. 279. 

inverted in-line 

Fig. 279. 

Ranger 6 cylinder in-line, inverted, air-cooled, aircraft engine. (Courtesy 

of Aviation.)

C/iapterXV 

COMPRESSION RATIO AND VALVE TIMING 

In an actual engine cylinder, the piston at top center 

does not touch the top of the cylinder. The space left near 

the top of the cylinder after the piston has reached top 

center may have any of a wide variety of shapes depending 

>) (b) ~(c) (d) (6) 

Fig. 280. Types of combustion chambers from "The Airplane and Its Engine'* by 

Chatfield, Taylor, and Ober. 

upon the engine design. Some of the more common shapes 

are shown in Fig. 280. 

Job 1 : 

Cylinder Volume 

The number of cylinders in aircraft engines ranges from 

2 for the Aeronca all the way up to 14 cylinders 

for certain 

Wright or Pratt and Whitney engines. 

For all practical purposes, all cylinders of a multicylinder 

engine may be considered identical. It was therefore con- 

224

Compression Ratio and Valve Timing 225 

sidered best to base the definitions and formulas in this job 

upon a consideration of one cylinder only, as shown in Fig. 

281. However, these same definitions and formulas will also 

hold true for the entire engine. 

B.C. 

Fig. 281 .Cylinder from Pratt and Whitney Wasp. (Courtesy of Aviation.) 

Definitions: 

1. Clearance volume is the volume of the space left above 

the piston when it is 

at top center. 

Note: This is sometimes called the volume of the combustion 

chamber. 

2. Displacement is the volume that the piston moves 

through from bottom center to top center. 

3. Total volume of 1 cylinder is equal to the displacement 

plus the clearance volume. 

Formula :V, 

^ Disp. + V c 

where V c means clearance volume. 

Disp. means displacement for one cylinder. 

V t 

means total volume of one cylinder. 


The displacement of a cylinder is 70 cu. in. and the clearance 

volume is 10 cu. in. Find the total volume of 1 cylinder.


Given: Disp. 

= 70 cu. in. 

Find: 

V t 

V c 

= 10 cu. in. 

Examples: 

V t 

= Disp. + V c 

V t 

= 70 + 10 

Vt = 80 cu. in. Ans. 

1. The displacement of one cylinder of a Kinner is 

98 cu. in. The volume above the piston at top center is 

24.5 cu. in. What is the total volume of one cylinder? 

2. Each cylinder of a Whirlwind engine has a displacement 

of 108 cu. in. and a clearance volume of 18 cu. in. 

What is the volume of one cylinder? 

3. The Whirlwind engine in Example 2 has 9 cylinders. 

What is the total displacement? What is the total volume 

of all cylinders? 

4. The total volume of each cylinder of an Axelson aircraft 

engine is 

105 cu. in. Find the clearance volume if the 

displacement for one cylinder is 

Job 2: Compression Ratio 

85.5 cu. in. 

The words compression ratio are now being used so much 

in trade literature, instruction manuals, and ordinary auto- 

Totat cyUnder volume 

Bottom 

center 

Fig. 282. The ratio of these two volumes is called the "compression ratio." 

mobile advertisements, that every mechanic ought to know 

what they mean.


It has been pointed out that the piston at top center does 

not touch the top of the cylinder. There is 

always a compression 

space left, the volume of which is called the 

clearance volume (see Fig. 282). 

Definition: 

Compression 

ratio is 

of one cylinder and its clearance volume. 

w 

Formula: C.R. = ~ 

Ve 

the ratio between the total volume 

where C.R. = compression ratio. 

Vt 

= total volume of one cylinder. 

V c 

= clearance volume of one cylinder. 

Here are some actual compression 

aircraft engines: 

ratios for various 

TABLE 14 

Engine 

Compression Ratio 

Jacobs L-5 6:1* 

. 15: 1 

Aeronca E-113-C. 5.4:1 

Pratt and Whitney Wasp Jr 6:1 

Ranger 6. . . . 6.5:1 

Guiberson Diesel * Pronounced "6 to 1." 

Notice that the compression ratio of the diesel engine is 

much higher than that of the others. Why? 


Find the compression ratio of the Aeronca E-113-C in which 

the total volume of one cylinder is 69.65 cu. in. and the clearance 

volume is 12.9 cu. in. 

Given: V t 

= 69.65 cu. in. 

= 12.9 cu. in. 

Find: 

Vc 

C.R. 

C.R. - 

C>R - 

c 

69.65 

~ 12^ 

C.R. = 5.4 

Ans.


Examples: 

1. The total volume of one cylinder of the water-cooled 

Allison V-1710-C6 is 171.0 cu. in. The volume of the compression 

chamber of one cylinder is 28.5 cu. in. WhaHs the 

compression ratio? 

2. The Jacobs L-4M radial engine has a clearance 

volume for one cylinder equal to 24.7 cu. in. Find the compression 

ratio if the total volume of one cylinder is 132.8 

cu. in. 

3. Find the compression ratio of the 4 cylinder Menasco 

Pirate, if the total volume of all 4 cylinders is 

443.6 cu. in. 

and the total volume of all 4 combustion chambers is 

80.6 cu. in. 

Job 3: How to Find the Clearance Volume 

The shape of the compression chamber above the piston 

at top center will depend upon the type of engine, the 

number of valves, spark plugs, etc. Yet there is a simple 

method of calculating its volume, if we know the compression 

ratio and the displacement. Do specifications give 

these facts ? 

Notice that the displacement for one cylinder must 

be calculated, since only total displacement is given in 

specifications. 

r .. 

, displacement 

Formula: V c 

= -7^-5 

N^.K. 

I 

where V c 

= clearance volume for one cylinder. 

C.R. = compression ratio. 


The displacement for one cylinder 

is 25 cu. in.; its 

compression 

ratio is 6:1. Find it clearance volume. 

Given: Disp. 

= 25 cu. in. 

C.R. = 6:1

Find: V c 

Disp. 


Check the answer. 

V * c f ^ 

C.R. - 1 

25 

V c 

= 

6 - 1 

V c 

= 5 cu. in. Ans. 

Examples: 

1. The displacement for one cylinder of a LeBlond 

engine is 54 cu. in.; its compression ratio is 5.5 to 1. Find 

the clearance volume of one cylinder. 

2. The compression ratio of the Franklin is 5.5:1, and 

the displacement for one cylinder is 37.5 cu. in. Find the 

volume of the compression chamber and the total volume 

of one cylinder. Check the answers. 

is 

3. The displacement for all 4 cylinders of a Lycoming 

144.5 cu. in. Find the clearance volume for one cylinder, 

if the compression ratio is 5.65 to 1. 

4. The Allison V water-cooled engine has a bore and 

stroke of 5^ by 6 in. Find the total volume of all 12 cylinders 

if the compression 

ratio is 6.00:1. 

Job 4: 

Valve Timing Diagrams 

The exact time at which the intake and exhaust valves 

open and close has been carefully set by the designer, so as 

to obtain the best possible operation of the engine. After 

the engifte has been running for some time, however, the 

valve timing will often be found to need adjustment. 

Failure to make such corrections will result in a serious loss 

of power and in eventual damage to the engine. 

Valve timing, therefore, is an essential part 

of the 

specifications of an engine, whether aircraft, automobile, 

marine, or any other kind. All valve timing checks and 

adjustments that the mechanic makes from time to time 

depend upon this information.


A. Intake. Many students are under the impression that 

the intake valve always opens just as the piston begins to 

move downward on the intake stroke. Although this may 

at first glance seem natural, it is 

very seldom correct for 

aircraft engines. 

Intake valve 

opens ^^ 

v 

before top center, 

Intake valve 

opens 

22B.T.C. 

Arrowshows^ 

direction of 

rotation ofthe 

crankshaft 

Intake valve 

doses 62'A.B.C} 

Bottom 

Bottom 

Fi 9 283*. . Fig. 283b. 

center 

center 

Valve timing data is given in degrees. For instance, the 

intake valve of the Khmer K-5 opens 22 before top center. 

This can be diagrammed as shown in Fig. 283 (a). 

that the direction of rotation of the crankshaft is 

Intake 

opens 

22B.T.C. 

the arrow* 

Notice 

given by 

In most aircraft engines, the 

intake valve does not close as soon as 

the piston reaches the bottom of its 

downward stroke, but remains open 

for a considerable length of time 

thereafter. 

The intake valve of the engine 

shown in Fig. 283(6) closes 82 after 

bottom center. This information 

can be put on the same diagram. 

The diagram in Fig. 284 shows the valve timing diagram 

for the intake stroke. 

These abbreviations are used:


Top center ........ T.C. After top center .... A.T.C. 

Bottom center ..... B.C. Before bottom center B.B.C. 

Before top center. . . B.T.C After bottom center. A.B.C. 

Examples: 

1-3. Draw the valve timing diagram for the following 

engines. Notice that data for the intake valve only is given 

here. 

Intake valve 

B. Exhaust. Complete valve timing information naturally 

gives data for both the intake and exhaust valve. For 

T.C. 

Fig. 285. 

example, for the Kinner K-5 engine, the intake valve opens 

22 B.T.C. and closes 82 A.B.C.; the exhaust valve opens 

68 B.B.C. and closes 36 A.T.C. 

Examine the valve timing diagram in Fig. 285 (a) for the 

exhaust stroke alone. The complete valve timing diagram 

is shown in Fig. 285 (Z>).

232 MatAemat/cs for the Aviation Trades 

Examples: 

1-5. Draw the timing diagram for the following engines : 

Figure 286 shows how the Instruction Book of the 

Axelson Engine Company gives the timing diagram for 

poinnAdwcedW&C. 

Inlet opens 6B.TC:"/ 

TC 

V Valves honfe overlap 

1 K 

~%J&:E*t


When the piston is at bottom center, the throw is at its 

farthest point away from the cylinder. The shaft has 

turned through an angle of 180 just for the downward 

movement of the piston from top center to bottom center. 

Bofhi 

cento 

Fig. 287. 

The intake valve of the Kinner K-5 opens 22 B.T.C., 

and closes 82 A.B.C. The intake valve of the Kinner, therefore, 

remains open 22 + 180 + 82 or a total of 284. The 

exhaust valve of the Kinner opens 68 B.B.C., and closes 

1C. 

idO 

B.C. 

Intake 

Fig. 288. 

B.C. 

Exhaust 

36 A.T.C. It is, therefore, open 68 + 180 + 36 or a 

total of 284 (see Fig. 288). 

Examples: 

1-3. Draw the valve timing diagrams for the following 

that each valve 

engines and find the number of degrees 

remains open:


Job 6: 

Valve Overlap 

From the specifications given in previous jobs, it may 

have been noticed that in most aircraft engines the intake 

valve opens before the exhaust valve closes. Of course, this 

means that some fuel will be wasted. However, the rush of 

gasoline from the intake manifold serves to drive out all 

previous exhaust vapor and 

^-V&f/ve over-fa. 

leave the mixture in the cylinder 

clean for the next stroke. This 

'Exhaust 

valve 

closes 

is 

very important in a highcompression 

engine, since an 

improper mixture might cause 

detonation or engine knock. 

Definition: 

Valve overlap 

is the length of 

l9 ' 

' 

time that both valves remain 

open at the same time. It is measured in degrees. 

In finding the valve overlap, it will only be necessary to 

consider when the exhaust valve closes and the intake 

valve opens as shown in Fig. 289. 


What is the valve overlap for the Kinner K-5 ? 

Given: Exhaust valve closes 36 A.T.C. 

Intake valve opens 22 B.T.C.


Examples: 

Find: 

Valve overlap 

Valve overlap 

= 22 + 36 = 58 Ans. 

1-3. Find the valve overlap for each of the following 

engines. First draw the valve timing diagram. Is there any 

overlap for the Packard engine? 


The specifications and performance curves (Fig. 290) for 

the Jacobs model L-6, 7 cylinder radial, 

air-cooled engine 

1300 1500 1700 1900 2100 2300 2500 

R.p.m. 

Fig. 290. 

Performance curves: Jacobs L-6 aircraft engine.


Fig. 291 . 

Jacobs L-6 radial air-cooled aircraft engine. 

(Fig. 291) follow. All the examples in this job will refer to 

these specifications: 

Name and model Jacobs L-6 

Type 

A.T.C. No 195 

Cylinders 7 

Bore 5 in. 

Stroke 

B.M.E.P 

Direct drive, air-cooled, radial 

5J in. 


R.p.m. at rated hp 2,100 

Compression ratio 6:1 

Specific oil consumption 

0.025 Ib. per hp.-hr. 

Specific fuel consumption 

0.45 Ib. per hp.-br. 

Valve timing information: 

Intake opens 18 B.T.C.; closes 65 A.B.C. 

Exhaust opens 58 B.B.C; closes 16 A.T.C. 

Crankshaft rotation, looking from rear of engine, clockwise


Examples: 

1. Find the area of 1 piston. Find the total piston area, 

2. What is the total displacement for all cylinders? 

3. Calculate the brake horsepower at 2,100 r.p.m. 

4. Complete these tables from the performance curves: 

5. Find the clearance volume for 1 cylinder. 

6. Find the total volume of 1 cylinder. 

7. Draw the valve timing diagram. 

8. How many degrees does each valve remain open ? 

9. What is the valve overlap in degrees ? 

10. How many gallons of gasoline would this engine 

consume operating at 2,100 r.p.m. for 1 hr. 35 min.?

Party 

REVIEW 

S39

Chapter XVI 

ONE HUNDRED SELECTED REVIEW 

EXAMPLES 

Fig. 292: 

1. Can you read the rule? Measure the distances in 

A 

H/> 

-+\B 

E\+ 

F 

Fig. 292. 

(a) AB (b) AD (c) EF (d) OF 

(e) GE (/) DK 

2. Add: 

() i + 1 + i + iV (b) i + f + f 

(c) 

much ? 

3. Which fraction in each group is the larger and how 

() iorff (6) fVor^ 

(c) -fa or i (rf) ^ 

or i 

4. Find the over-all dimensions of the piece in Fig. 293. 

Fi S . 293. 

241


5. Find the over-all dimensions of the tie rod terminal 


ftf o// jH 

--g- blr 

J 

Fig. 294. 

6. Find the missing dimension in Fig. 295. 

ft* 

'%- 

Fig. 295. 

7. Find the missing 

dimension of the beam shown in 

Fig. 296. t-0.625 -: 

c- -2.125"- * 

Fig. 296. 

8. Multiply: 

(a) 3.1416 X 25 (6) 6.250 X 0.375 (c) 6.055 X 1.385 

9. Multiply: 

(a) f X f (6) 12^ X f (c) 3^ X

One Hundred Selected Review Examples 243 

10. Find the weight of 35 ft. of round steel rod, 

if the 

weight is 1.043 Ib. per ft. of length. 

11. If 1-in. round stainless steel bar weighs 2.934 Ib. per 

ft. 

of length, find the weight of 7 bars, each 18 ft. long. 

12. Divide: 

(a) 2i by 4 (6) 4^ by f (c) 12f by 

13. Divide. Obtain answers to the nearest hundredth, 

(a) 43.625 by 9 (6) 2.03726 by 3.14 (c) 0.625 by 0.032 

14. The strip of metal in Fig. 297 is to have the centers 

of 7 holes equally spaced. Find the distance between centers 

to the nearest 64th of an inch. 

(J) 

20'^ 

Fig. 297. 

16. How many round pieces |- in. in diameter can be 

" punched" from a strip of steel 36 in. long, allowing iV in. 

between punchings (see Fig. 298) ? 

^f 

Stock: '/Q 

thick, /"wide 

Fig. 298. The steel strip is 36 in. long. 

16. a. What is the weight of the unpunched strip in 

Fig. 298? b. What is the total weight of all the round punchings? 

c. What is the weight of the punched strip ? 

17. Find the area of each figure in Fig. 299.

: 

-J./J0" 


fa) 

Fig. 299. 

(b) 

18. Find the perimeter of each figure in Fig. 299. 

19. Find the area and circumference of a circle whose 

diameter is 4ijr in. 

20. Find the area in square inches of each figure in 

Fig. 300. 

(a.) 

Fi 3 . 300. (6) 

21. Find the area of the irregular flat surface shown in 

Fig. 801. 

-H V-0.500" 

Fig. 301. 

22. Calculate the area of the cutout portions of Fig. 302. 

Fig. 302. 

(b


23. Express answers to the nearest 10th: 

(6) VlS.374 (c) V0.9378 

24. What is the length of the side of a square whose area 

is 396.255 sq. in.? 

25. Find the diameter of a piston whose face area is 

30.25 sq. in. 

26. Find the radius of circle whose area is 3.1416 sq. ft. 

27. A rectangular field whose area is 576 sq. yd. has a 

length of 275 ft. What is its width ? 

28. For mass production of aircraft, a modern brick and 

steel structure was recently suggested comprising the 

following sections: 

Section 

Dimension, Ft. 

Manufacturing 600 by 1,400 

Engineering 100 by 900 

Office 50 by 400 

Truck garage 120 by 150 

Boiler house 100 by 100 

Oil house 75 by 150 

Flight hangar 200 by 200 

a. Calculate the amount of space in square feet assigned 

to each section. 

6. Find the total amount of floor space. 

29. Find the volume in cubic inches, of each solid in 

Fig. 303. Fig. 303. 

30. How many gallons of oil can be stored in a circular 

tank, if the diameter of its base is 25 ft. and its height is 

12 feet?

246 Mathematics for t/)e Aviation Trades 

31. A circular boiler, 8 ft. long and 4 ft. 6 in. in diameter, 

is completely filled with gasoline. What is the weight of the 

gasoline ? 

32. What is the weight of 50 oak 

beams each 2 by 4 in. by 12 ft. 

long? 

33. Calculate the weight of 5,000 of 

the steel items in Fig. 304. 

34. Find the weight of one dozen 

. 304. 

copper plates cut according to the 

dimensions shown in Fig. 305. 

12 Pieces 

" 

'/ 

4 thick 

V- 15" 4 22- 4 

Fi g . 305. 

35. Calculate the weight of 144 steel pins 

Fig. 306. 

as shown in 

36. How many board feet are there in a piece of lumber 

2 in. thick, 9 in. wide, and 12 ft. long? 

37. How many board feet of lumber would be needed to 

build the platform shown in Fig. 307 ? 

38. What would be the cost of this bill of material ?


39. Find the number of board feet, the cost, and the 

weight of 15 spruce planks each f by 12 in. by 10 ft., 

price is $.18 per board foot. 

if the 

40. Calculate the number of board feet needed to construct 

the open box shown in Fig. 308, if 1-in. white pine is 

used throughout. 

41. (a) What is the weight of the box (Fig. 308) ? (6) 

What would be the weight of a similar steel box? 

42. What weight of concrete would the box (Fig, 308) 

contain when filled? Concrete weighs 150 Ib. per cu. ft. 

43. The graph shown in Fig. 309 appeared in the 

Nov. 15, 1940, issue of the Civil Aeronautics Journal. 

Notice how much information is 

given 

space. 

12.0 

UNITED STATES Aiu TRANSPORTATION 

REVENUE MILES FLOWN 

in this small 

10.5 

9.0 

o 

-7.5 

6.0 

1940 

7 

~7 // 

4.5 

Join. Feb. Mar. Apr. May June July Aug. Sept, Oct. Nov. Dec. 

Fig. 309. (Courtesy of Civil Aeronautics Journal.)


a. What is the worst month of every year shown in the 

graph 

Why? 

as far as "revenue miles flown" is concerned? 

6. How many revenue miles were flown in March, 1938? 

In March, 1939? In March, 1940? 

44. Complete a table of data showing 

revenue miles flown in 1939 (see Fig. 309). 

the number of 

46. The following table shows how four major airlines 

compare with respect to the number of paid passengers 

carried during September, 1940. 

Operator 

Passengers 

American Airlines 93,876 

Eastern Airlines 33,878 

T.W.A 35,701 

United Air Lines 48,836 

Draw a bar graph of this information. 

46. Find the over-all length of the fitting shown in 

Fig. 310. 

Section A-A 

Fi 3 310. 1/8-in. . cold-rolled, S.A.E. 1025, 2 holes drilled 3/16 in. diameter. 

310)? 

47. Make a full-scale drawing of the fitting in Fig. 310. 

48. Find the top surface area of the fitting in Fig. 310. 

49. What is the volume of one fitting? 

60. What is the weight of 1,000 such items? 

51. What is the tensile strength at section AA (Fig. 

62. What is the bearing strength at AA (Fig. 310)? 

53. Complete a table of data in inches to the nearest 

64th for a 30-in. chord of airfoil section N.A.C.A. 22 from 

the data shown in Fig. 311,

One Hundred Se/ectec/ Review Examples 249 

Fig. 311. Airfoil section: N.A.C.A. 22. 

N.A.C.A. 22 

54. Draw the nosepiece (0-15 per cent) from the data 

obtained in Example 53, and construct a solid wood nosepiece 

from the drawing. 

56. What is the thickness in inches at each station of the 

complete 

airfoil for a 30-in. chord ? 

56. Draw the tail section (75-100 per cent) 

chord length of the N.A.C.A. 22. 

for a 5-ft 

57. Make a table of data to fit the airfoil shown in 

Fig. 312, accurate to the nearest 64th.


Fig. 312. 

Airfoil section. 

58. Design an original airfoil section on graph paper 

and complete a table of data to go with it, 

59. What is the difference between the airfoil section 

in Example 58 and those found in N.A.C.A. references? 

60. Complete a table of data, accurate to the nearest 

tenth of an inch, for a20-in. chord of airfoil section N.A.C.A. 

4412 (see Fig. 313). 



20 

20 40 60 80 


Fig. 313. Airfoil section: N.A.C.A. 4412 is used on the Luscombe Model 50 

two-place monoplane. 

61. Draw a nosepiece (0-15 per cent) for a 4-ft. chord of 

airfoil section N.A.C.A. 4412 (see Fig. 313).


62. What is the thickness at each station of the nosepiece 

drawn in Example 61 ? Check the answers by actual 

measurement or by calculation from the data. 

63. Find the useful load of the airplane (Fig. 314). 

Fig. 31 4. Lockheed Lodestar twin-engine transport. (Courtesy of Aviation.) 

LOCKHEED LODESTAR 

Weight, empty 

12,045 Ib. 

Gross weight 

17,500 Ib. 

Engines 

2 Pratt and Whitney, 1200 hp. each 

Wing area 

551 sq. ft. 

Wing span 65 ft. 6 in. 

64. What is the wing loading? What is the power 

loading? 

66. What is the mean chord of the wing? 

66. Find the aspect ratio of the wing. 

67. Estimate the dihedral angle of the wing from Fig. 

314. 

68. Estimate the angle of sweepback? 

69. What per cent of the gross weight 

is the useful load ? 

70. This airplane (Fig. 314) carries 644 gal. of gasoline, 

and at cruising speed each engine consumes 27.5 gal. per hr* 

Approximately how long can it stay aloft? 

would use to find: 

71. What is the formula you 

a. Area of a piston? 

6. Displacement? 

c. Number of power strokes per minute? 

d. Brake horsepower of an engine?


e. Fuel consumption? 

/. Compression ratio? 

g. Clearance volume? 

72. Complete the following table. Express answers to 

the nearest hundredth. 

FIVE CYLINDER KINNER AIRCRAFT ENGINES 

73. Represent the results from Example 72 graphically, 

using any appropriate type of graph. The Kinner K-5 is 


Pis. 31 5. Kinner K-5, 5 cylinder, air-cooled, radial aircraft engine.

One Hundred Se/ectec/ Review Examples 253 

PERFORMANCE. 

CURVES 

R.RM. 

Fig. 316. 

Lycoming seared 75-hp. engine. 

^^^î^'O^ vi'J ,,'.:',.. ','rtteirJ, 

.' 

Fig. 317. Lycoming geared 75-hp., 4 cylinder opposed, air-cooled, aircraft 

engine. (Courtesy of Aviation.)


The specification and performance curves (Fig. 316) are 

for the Lycoming geared 75-hp. engine shown in Fig. 317. 

Number of cylinders 4 

Bore 

3.625 in. 

Stroke 

Engine r.p.m 

B.M.E.P 

8.50 in. 

8,200 at rated horsepower 

1$8 lb. per sq. in. 

Compression ratio 0.5: 1 

Weight, dry 

181 lb 

Specific fuel consumption 

0.50 Ib./b.hp./hr. 

Specific oil consumption 

0.010 Ib./b.hp./hr. 

Valve Timing Information 

Intake valve opens 20 B.T.C.; closes 65 A. B.C. 

Exhaust valve opens 65 B.B.C.; closes 20 A.T.C. 

74. What is the total piston area? 

75. What is the total displacement? 

76. Calculate the rated horsepower of this engine. 

Is it 

exactly 75 hp.? Why? 

77. What is the weight per horsepower of the Lycoming ? 

78. How many gallons of gasoline would be consumed 

if the Lycoming operated for 2 hr. 15 min. at 75 hp.? 

79. How many quarts of oil would be consumed during 

this interval? 

80. Complete the following 

performance curves: 

table of data from the 

81. On what three factors does the bend allowance 

depend ?


82. Calculate the bend allowance for the fitting shown 

in Fig. 318. +\ \+0.032" 

/&' 

Fis. 318. Ansle of bend 90. 

83. Find the over-all or developed length of the fitting 

in Fig. 318. 

84. Complete the following table: 

BEND ALLOWANCE CHART: 90 ANGLE OF BEND 

(All dimensions are in inches) 

0.049 

0.035 

0.028 

Use the above table to help solve the examples that follow. 

85. Find the developed length of the fitting shown in 

Fig. 319. s- 319. An 9 le of bend 90.

256 Mat/iemat/cs for the Aviation Trades 

86. Calculate the developed length of the part shown in 

Fig. 320. 0.028- 

,/L 

Fig. 320. Two bends, each 90. 

87. What is the strength in tension of a dural rod whose 

diameter is 0.125 in.? 

88. Find the strength in compression parallel to the 

grain of an oak beam whose cross section is 2-g- by 3f in. 

89. What would be the weight of the beam in Example 

88 if it were 7 ft. 

long? 

90. Calculate the strength in shear of a ^V-in. copper 

rivet. 

91. What is the strength in shear of the lap joint shown 

in Fig. 321 ? 

o 

Fig. 321. Two 1/16-in. S.A.E. X-4130 rivets. 

92. What is the strength in bearing of a 0.238-in. dural 

plate with a ^-in. rivet hole? 

93. Find the strength in tension and bearing of the 

cast-iron lug shown in Fig. 322. 

Fig. 322. 

94. What is the diameter of a L.C. steel wire that can 

hold a maximum load of 1,500 lb.?


95. A dural tube has an outside diameter of l in. and a 

wall thickness of 0.083 in. 

a. What is the inside diameter? 

b. What is the cross-sectional area? 

weigh ? 

96. What would 100 ft. of the tubing in Example 95 

97. If no bending or buckling took place, what would 

be the maximum compressive strength that a 22 gage 

(0.028 in.) S.A.E. 1015 tube could develop, if its inside 

diameter were 0.930 in.? 

98. Find the strength in tension of the riveted strap 


Fig. 323. Lap joint, dural straps. Two 1 /8 in., 2S rivets, driven cold. 

99. What is the strength in shear of the joint in Fig. 323 ? 

100. What is the strength in shear of the butt joint shown 

in Fig. 324? 

-1X9 

Fig. 324. All rivets 3/64 in. 17 S-T, driven hot.

APPENDIX 

TABLES AND FORMULAS 

259

Tables of Measure 

TABLE 1. 

LENGTH 

12 inches (in. or ") 

= 1 foot (ft. or ') 

3 feet = 1 yard (yd.) 

5j yards 

= 1 rod 

5,280 feet = 1 mile (mi.) 

1 meter (m.) 

= 39 in. (approximately) 

TABLE 2. 

AREA 

144 square inches (sq. in.) = 1 

square foot (sq. ft.) 

9 square feet = 1 

square yard (sq. yd.) 

4840 square yards 

= 1 acre 

640 acres = 1 square mile (sq. mi.) 

TABLE 3. 

VOLUME 

1728 cubic inches (cu. in.) 

= 1 cubic foot (cu. ft.) 

27 cubic feet = 1 cubic yard (cu. yd.) 

1 liter = 1 

quart (approximately) 

TABLE 4. 

ANGLE MEASURE 

60 seconds (sec. or ") = 1 minute (min. or ') 

60 minutes = 1 degree () 

90 degrees 

= 1 right angle 

180 degrees 

= 1 straight angle 

360 degrees 

= 1 complete rotation, or circle 

TABLE 5. COMMON WEIGHT 

16 ounces (oz.) 

= 1 pound (Ib.) 

2000 pounds = 1 ton (T.) 

1 kilogram 

= 2.2 pounds (Ib.) (approximately) 

261


Formulas 

FROM PART I, 

A REVIEW OF FUNDAMENTALS 

FROM PART II, THE AIRPLANE AND ITS WING 

Wing area = span X chord 

Mean chord 

wing area 

span 

A L ,.' 

Aspect ratio 

sPan 

= -f- -j 

chord 

Gross weight 

= empty weight + useful load 

w 

. 

Wing loading = 5 ^~ 

n , 

, j. gross weight 

wing area 

.. gross weight 

Power loading 

= -,horsepower 

FROM PART III, 

MATHEMATICS OP MATERIALS 

Tensile strength 

= area X ultimate tensile strength 

Compressive strength = area X ultimate compressive strength 

Shear strength 

= area X ultimate shear strength 

Cross-sectional area 

strength required 

ultimate strength

where R 

Appendix 

263 

Bend allowance = (0.01743 X R + 0.0078 X T) X N* 

radius of bend. 

T = thickness of the metal. 

N number of degrees in the angle of bend. 

FROM PART IV, AIRCRAFT ENGINE MATHEMATICS 

Displacement = piston area X stroke 

Number of power strokes per min. = -^ 

Indicated hp. 

= brake hp. + friction hp. 


n - 

W.np. , 

X cylinders 

38,000 

X F X D X r.p.m. 

B.hp. - 

(using 

33,000 

the Prony brake) 

Compression ratio total cylinder volume 

clearance volume 

Decimal Equivalents 

KOI 5625 

-.03125 

K046875 

-.0625 

K078 1 25 

-.09375 

Kl 09375 

-.125 

KI40625 

-.15625 

KI7I875 

-.1875 

H203I25 

-.21875 

KgUTS 

K 26 5625 

-.28125 

K296875 

-.3125 

K328I25 

-.34375 

K359375 

^.375 

K390625 

-.40625 

K42I675 

K453I25 

-.46875 

K484375 

-.5

INDEX 

Accuracy of measurement, 5-8, 32 

Addition of decimals, 22 

of nonruler fractions, 40 

of ruler fractions, 12 

Aircraft engine, 

191-237 

performance curves, 221, 235, 253 

Airfoil section, 130-150 

with data, in inches, 131 

per cent of chord, 135 

with negative numbers, 144 

nosepiece of, 139 

tailsection of, 139 

thickness of, 142 

Airplane wing, 111-150 

area of, 115 

aspect ratio of, 117 

chord, 115 

loading, 124 

span of, 115 

Angles, 80-89, 93, 182 

in aviation, 86 

how bisected, 88 

how drawn, 82, 93 

how measured, 84 

units of measure of, 84 

Area, 47-69 

of airplane wing, 1 13 

of circle, 61 

cross-sectional, required, 164 

formulas for, 261 

of piston, 195 

of rectangle, 48 

of square, 58 

of trapezoid, 66 

of triangle, 64 

265 

Area, units of, 47, 261 

Aspect ratio, 117 

Bar graph, 98-101 

Bearing, 162-164 

B 

strength, table of, 162 

Bend allowance, 181-190 

Board feet, 76-78 

Broken-line graph, 103-105 

Camber, upper and lower, 131-135 

Chord mean, 116 

per cent of, 135 

Circle, area of, 61 

circumference of, 43 

Clearance volume, 228, 229 

Compression, 157-160 


Compression ratio, 226-228 

Construction, 88-97 

of angle bisector, 88 

of equal angle, 93 

of line bisector, 89 

of line into equal parts, 96 

of parallel line, 94 

of perpendicular, 91 

Curved-line graph, 105, 106 

Cylinder volume, 224-226 

Decimals, 20-36 

D 

checking dimensions with, 22


Decimals, division of, 26 

to fractions, 27 

multiplication of, 24 

square root of, 56 

Displacement, 197-199 

Equivalents, chart of decimal, 29 

E 

Mean effective pressure, 203-205 

Measuring, accuracy of, 5 

length, 87-45 

with protractor, 81 

with steel rule, 3 

Micrometer caliper, 20 

Mixed numbers, 10-12 

Multiplication, of decimals, 24-26 

of fractions, 15 

Fittings, 169-171 

Formulas, 260-262 

Fractions, 8-18, 40-42 

addition of, 12, 40 

changing to decimals, 27 

division of, 16 

multiplication of, 15 

reducing to lowest terms, 8 

subtraction of, 18 

Fuel and oil consumption, 212-220 

gallons and cost, 215 

specific, 213, 217 

Parallel lines, 94 

Pay load, 121-123 

Perimeter, 39 

Perpendicular, 91, 92 

Pictograph, 101-103 

Piston area, 195-197 

Power loading, 126 

Power strokes, 199, 204 

Prony brake, 208-210 

Protractor, 80-82 

R 

Graphs, 98-108 

bar, 98 

broken-line, 108 

curved-line, 105 

pictograph, 101 H 

Horsepower, 193-210 

formula for, 206 

types of, 201 

Horsepower-hours, 212, 218 

Improper fractions, 1 1 

M 

Materials, strength of, 153-179 

weight of, 73-76 

Rectangle, 48-51 

Review, selected examples, 241-257 

Review tests, 18, 34, 45, 68, 78, 97, 

108, 127, 148, 166, 179, 189, 210, 

220, 235 

Riveted joints, 177-179 

Rivets, 175-179 

strength of, 175, 177 

types of, 175, 176 

Shear, 160-162 


Span of airplane wing, 115 

Specific consumption, 218-217 

of fuel, 218 

of oil, 217 

Square, 58-61 

Square root, 54-56 

of decimals, 56 

of whole numbers, 55

Squaring a number, 52-54 

Steel rule, 3-8 

Strength of materials, 153-179 

bearing, 162 

compression, 157 

safe working, 168 

table of, 169 

shear, 160 

tenslon ' 

154 

Tables of measure, 259 

Tension, 154-157 

Index 267 

Useful load > 12 

U 

y 

Valve timin g- 229 ~ 235 

diag 'ams ' 

229 

overlap 234 

Volume, of aircraft-engine cylinder, 224 

of clearance, 228 

formula for, 71, 72 

units of, 70 

and weight, 70-78 

strength, table of, 155 ,, . 

r , . . . 

5 

Weight of an airplane, 113-127 

Thickness gage, 23 

formula ^ ?4 

Tolerance and limits, 32-34 

gposg and empty> 119 

Trapezoid, 66-68 of mat erials, 73-76 

Triangle, 64-66 table of, 74 

Tubing, 171-175 Wing loading, 124-126 

W

MATHEMATICS - SprueMaster

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